Derivatives and Its Application

Transcription

1 Chapter 4 Derivatives an Its Application Contents 4.1 Definition an Properties of erivatives; basic rules; chain rules 3 4. Derivatives of Inverse Functions; Inverse Trigonometric Functions; Hyperbolic Functions an Inverse Hyperbolic Functions Implicit Differentiation; Higher Orer Derivatives Implicit Differentiation Higher Derivatives Applications of Derivative Ineterminate Forms an L Hopital s Rule Ineterminate Forms Exercise 33 Derivative measures the steepness of the graph of a function at some particular point on the graph. Thus, the erivative is a slope that is a ratio of change in the value of the function to change in the inepenent variable. If the inepenent variable is time, we often think of this ratio as a rate of change. If we zoom in on the graph of the function at some point that the function looks almost like a straight line, the erivative at that point is the slope of the line. This is the same as saying that the erivative is the slope of the tangent line to the graph of the function at the given point. The slope of a secant line which is connecting two points on a graph approaches the erivative when the interval between the points shrinks own to zero. The erivative itself is also a function that varies from place to place. For example, the velocity of a car may change from moment to moment as the river spees up or slows own. 4.1 Definition an Properties of erivatives; basic rules; chain rules Definition of Derivative Definition The tangent line to the curve y f (x) at the point P (a, f (a))is the line through with slope f (x) f (a) m x a x a provie that this it exists. Example 4.1 Fin an equation of the tangent line to the parabola y x at the point (1,1). Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

2 Derivatives an Its Application Solution: Here we have a 1 an f (x) x, so the slope is m f (x) f (1) x 1 x 1 m x 1 x 1 x 1 m (x 1)(x + 1) x 1 x 1 m (x + 1) x 1 m m Using the point-slope form of the equation of a line, we fin that an equation of the tangent line at (1,1) is y 1 (x 1) or y x 1. Definition 4.1. The erivative of a function f at any variable x, enote by f (x), is given by Provie the it exists. f (x) h 0 f (x + h) f (h) h Definition If we write x a + h, then h x a an approaches 0 if an only if x approaches a. Therefore, an equivalent way of stating the efinition of the erivative, as we saw in fining tangent lines, is f f (x) f (a) (a) x a x a Provie the it exists. The tangent line to the curve y f (x) at (a, f (a)) is the line through (a, f (a)) whose slope is equal to f (a), the erivative of f at a. Example 4. Fin the erivative of the function f (x) x 8x + 9 at the number a. Solution: From the above efinition we have f (a) h 0 f (a + h) f (a) h ((a + h) 8(a + h) + 9) (a 8a + 9) (a + ah + h 8a 8h + 9) (a 8a + 9) ah + h 8h (a + h 8) h 0 a 8 Example 4.3 If f (x) x 1, fin the erivative of f. State the omain of f. Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

3 4.1 Definition an Properties of erivatives; basic rules; chain rules 3 Solution: From the efinition of erivatives f f (x + h) f (x) (x) x + h 1 x 1 x + h 1 x 1 x + h 1 + x 1 x + h 1 + x 1 (x + h 1) (x 1) ( x + h 1 + x 1) h ( x + h 1 + x 1) 1 h 0 x + h 1 + x 1 1 x 1 + x 1 1 x 1 We see that f (x) exists if x > 1, so the omain of f is (1, ). If we use the traitional notation y f (x) to inicate that the inepenent variable is x an the epenent variable is y, then some common alternative notations for the erivative are as follows f (x) y y x f x x f (x) D f (x) D x f (x) The symbols D an x (Leibniz notation) are calle ifferentiation operators because they inicate the operation of ifferentiation, which is the process of calculating a erivative. If we want to inicate the value of a erivative x y in Leibniz notation at a specific number, we use the notation x xa, which is a synonym for f (a). Remark A function is ifferentiable at a if f (a) exists. It is ifferentiable on an open interval (a,b) or (a, ) or (,a) or (, ) if it is ifferentiable at every number in the interval. Example 4.4 Where is the function f (x) x ifferentiable? Solution: If x > 0, then x x an we can choose small enough that x + h > 0 an hence x + h x + h. Therefore, for x > 0 we have f (x) x + h x (x + h) x h 1 h 0 1 h 0 h an so f is ifferentiable for any x > 0. Similarly, for x < 0 we have x x an h can be chosen small Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

4 4 Derivatives an Its Application enough that x + h < 0 an so x + h (x + h). Therefore, for x < 0, an so f is ifferentiable for any x < 0. For x 0 we have to investigate f (x) x + h x (x + h) ( x) h 1 h 0 1 Let s compute the left an right its separately: h 0 h f (0) h 0 f (0 + h) f (x) h h h 0 h h if it exist h h 0 + h h h 1 an h h 0 h h h 1 Since these its are ifferent, f (0) oes not exist. Thus, f is ifferentiable at all x except 0. If the graph of a function f has a corner, curse or iscontinuity point at a point a, then the function o not ifferentiable at those point. Theorem 4.1. If f is ifferentiable at a, then f is continuous at a. But the converse of the above statement may not be true. Properties of Derivative If f (x) an g(x) are ifferentiable function an c is some constant numbers, then 1. Constant Function Rule x c 0. Power Function Rule x xn nx n 1 where n is any real number. 3. Constant Multiple Rule x (c f (x)) c x f (x) 4. Sum or Difference Rule x ( f (x) ± g(x)) x f (x) ± x g(x) 5. Prouct Rule x ( f (x) g(x)) f (x) x g(x) + ( x f (x))g(x) 6. Quotient Rule x ( f (x) g(x) ) ( x f (x))g(x) f (x) x g(x) (g(x)) provie g(x) 0 Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

5 4.1 Definition an Properties of erivatives; basic rules; chain rules 5 Example 4.5 If f (x) x 6, then f (x) 6x 5 by using power function rule. Example 4.6 Fin f (x) at x 1, where f (x) x 7 x x3 + 0x + 1. Solution: First we fin the erivatives of the function at an arbitrary x. f (x) x (x7 x x3 + 0x + 1) (7x 6 ) (5x 4 ) + 3 (3x ) x 6 5 x x + 0 Then we evaluate the erivatives at x 1 f ( 1) 14( 1) 6 5 ( 1) ( 1) Example 4.7 Fin f (x), where f (x) 9x7 x +1. Solution: If g(x) 9x 7 an h(x) x + 1, then f g h, g (x) 63x 6 an h (x) x. Therefore f (x) g (x)h(x) g(x)h (x) (h(x)) (63x6 )(x + 1) (9x 7 )(x) (x + 1) 45x8 + 63x 6 (x + 1) Rules of Derivatives 1. Exponential Functions The Derivative of the exponential function f (x) a x is f (x) a x lna. Definition Definition of the Number e e e is the number such that h 1 h 0 h 1, where e Derivative of the Natural Exponential Function 3. Derivatives of Logarithmic Functions x (ex ) e x 4. Derivatives of Trigonometric Functions (a) x sinx cosx (b) x cosx sinx (c) x tanx sec x () x cscx cscxcotx (e) x secx secxtanx (f) x cotx csc x x (log a x) 1 xlna an x (lnx) 1 x Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

6 6 Derivatives an Its Application Consier f (x) sinx. From the efinition of the erivative, we have f (x) f (x + h) f (x) sin(x + h) sinx sinxcosh + cosxsinh sinx sinxcosh sinx ( + cosx( sinh h )) sinxcosh sinx + cosx( sinh ) cosh 1 sinh sinx + cosx (sinx)(0) + (cosx)(1) cosx In general, the erivatives of the remaining functions can also be foun easily using the efinition an rules of erivatives. 1 tanx secx. Example 4.8 Differentiate f (x) Solution: Here f (x), then we have secx 1 tanx f (x) x ( secx ) 1 tanx ( x secx)(1 tanx) secx x (1 tanx) (1 tanx) secxtanx(1 tanx) secx( sec x) (1 tanx) secxtanx(1 tanx) + sec3 x (1 tanx) secx(tanx tan x tan x) 1 tanx + tan x secx(tanx + 1) sec x tanx Chain Rule Definition If h an g are both ifferentiable an f hog is the composition function efine by (hog)(x) h(g(x)), then f is ifferentiable an f is given by the prouct f (x) h (g(x))g (x). In Leibniz notation, if y f (u) an u g(x) are both ifferentiable functions, then y x y u u x The power rule combine with the chain Rule If n is any real number an u h(x) is ifferentiable, then Alternatively, x (h(x)) n n(h(x)) n 1 h (x). x (un n 1 u ) nu x Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

7 4.1 Definition an Properties of erivatives; basic rules; chain rules 7 Example 4.9 Fin f (x) if f (x) x + 3. Solution: f can be expresse as f (x) (hog)(x), where h(u) u an g(x) x + 3. Since We have f (u) 1 u 1 1 u an g (x) x f (x) h (g(x))g (x) 1 x + 3 x x x + 3 Example 4.10 Fin the erivative of f (x) (x x 3 ) 11. Solution: f (x) 11(x x 3 ) 1 x (x x3 ) 11(x x 3 ) 1 (1 6x ) 11(1 6x )(x x 3 ) 1 ( x )(x x 3 ) 1 (66x 11)(x x 3 ) 1 Example 4.11 Fin the erivative of f (x) sin 8 x. Solution: From the efinition of chain rule. We let u sinx. Then f (x) u 8, so it follows that f (x) x f (x) 8u 7 cosx 8sin 7 xcosx Example 4.1 Fin the erivative of y x sinx. Solution: Since y x sinx e lnxsinx e sinxlnx Thus, y y x x (esinxlnx ) e sinxlnx x (sinxlnx) x sinx x (sinxlnx) x sinx ( x (sinx)lnx + sinx x lnx) x sinx (cosxlnx + sinx 1 x ) x sinx (cosxlnx + sinx x ) Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

8 8 Derivatives an Its Application Example 4.13 Fin the erivative of y x xx. Solution: Since y x xx e lnxxx e xx lnx e elnxx lnx e exlnx lnx Thus, y y x x (eexlnx lnx ) e exlnx lnx x (exlnx lnx) x xx x (exlnx )lnx + e xlnx x (lnx)) x xx (lnx x exlnx ) + e xlnx 1 x ) x xx (e xlnx lnx x (xlnx)lnx + exlnx 1 x ) x xx (x x (lnx( x x lnx + x x lnx)) + xx 1 x ) x xx (x x ((lnx(lnx + 1) + 1 x ))) 4. Derivatives of Inverse Functions; Inverse Trigonometric Functions; Hyperbolic Functions an Inverse Hyperbolic Functions Derivatives of Inverse Functions Theorem 4..1 Suppose that f has an inverse an is continuous on an open interval I containing a. Assume also that f (a) exist, f (a) 0, an f (a) c. Then ( f 1 ) (c) exists, an Example 4.14 Let f (x) 8x 3 1 ( f 1 ) (c) 1 f (a) 1. Fin ( f 1 ) (7) by using the fact that f 1 (x) 1 (x + 1) Fin ( f 1 ) (7) by using the fact that f (1) 7. Solution: 1. The formula for f 1 appeare in the above example. Differentiating f 1, we fin that ( f 1 ) (x) 1 6 (x + 1) 3 So that. Since f (1) 7 an f (x) 4x, ( f 1 ) (x) 1 6 (7 + 1) ( f 1 ) (x) 1 f (1) 1 4 This is the same answer that we fin in the above part. Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

9 4. Derivatives of Inverse Functions; Inverse Trigonometric Functions; Hyperbolic Functions an Inverse Hyperbolic Functions 9 Example 4.15 Fin ( f 1 ) ( ), where f (x) x 7 + 8x 3 + 4x. Solution: First, we have to fin the value of a for which f (a). But f (a). So, a 0 an since f (x) 7x 6 + 4x + 4, it follows that f (0) 4. Therefore, ( f 1 ) ( ) 1 f (0) 1 4 Example 4.16 Let y x 5 + x. Fin x x y an y y 3. Solution: The function y has an inverse because its erivative, 5x 4 +, is a positive function. So x y 1 y x 1 5x 4 + Since y x 5 + x, it follows that y 3 for x 1. We conclue that x y 1 y 3 5( 1) Derivatives of Inverse Trigonometric Functions 1. x sin 1 x 1 1 x. x cos 1 x 1 1 x 3. x tan 1 x 1 1+x 4. x csc 1 x 1 5. x sec 1 x 1 x x x 1 x 1 6. x cot 1 x 1 x +1 Example 4.17 Show that x sin 1 x 1 1 x. Solution: Let y sin 1 x, then siny x, where π y π. Differentiating siny x respect to a variable y, then we have Now, cosy 0, since π y π. So, x y y x cosy cosy 1 cosy 1 sin y 1 x Therefore, x y 1 1 x. Example 4.18 Differentiate y 1 sin 1 x. Solution: Here y x x (sin 1 x) 1 (sin 1 x) x (sin 1 x) 1 (sin 1 x) 1 x Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

10 10 Derivatives an Its Application Hyperbolic Functions Definition of the Hyperbolic Functions 1. sinhx ex e x. coshx ex +e x 3. tanhx sinhx coshx 4. cschx 1 sinhx 5. sechx 1 coshx 6. cothx coshx sinhx Remark 4.. sinh has omain R an range R, while cosh has omain R an range [1, ). Hyperbolic Ientities 1. sinh( x) sinhx. cosh( x) cosh x 3. cosh x sinh x tanh x sech x 5. sinh(x + y) sinhxcoshy + coshxsinhy 6. cosh(x + y) coshxcoshy + sinhxsinhy Example 4.19 Prove that cosh x sinh x 1. Solution: From the efinition of hyperbolic function cosh x sinh x ( ex + e x ) ( ex e x ) ex + + e x ex + e x Inverse Hyperbolic Functions Inverse hyperbolic functions can be expresse as y sinh 1 x sinhy x y cosh 1 x coshy x an y 0 y tanh 1 x tanhy x Inverse hyperbolic functions can also be expresse in terms of logarithms. In particular, we have 1. sinh 1 x ln(x + x + 1),x R. cosh 1 x ln(x + x 1),x 1 3. tanh 1 x 1 1+x ln( 1 x ), 1 < x < 1 Example 4.0 Show that sinh 1 x ln(x + x + 1),x R. Solution: Let y sinh 1 x. Then, e y x e y 0 x sinh y ey e y Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

11 4. Derivatives of Inverse Functions; Inverse Trigonometric Functions; Hyperbolic Functions an Inverse Hyperbolic Functions 11 multiplying by e y, we have e y xe y 1 0 (e y ) x(e y ) 1 0 e y x ± 4x + 4 x ± x + 1 Note that e y > 0, but x x + 1 < 0. Thus, the minus sign will be ignore an we have e y x+ x + 1. Therefore, sinh 1 x ln(x + x + 1) since y sinh 1 x an the omain is the set of all real number. Example 4.1 Show that cosh 1 x ln(x + x 1),x 1. Solution: Let y cosh 1 x. Then, multiplying by e y, we have e y x + e y 0 x cosh y ey + e y e y xe y (e y ) x(e y ) e y x ± 4x 4 x ± x 1 y ln(x ± x 1) Note that y > 0, but ln(x x 1) < 0. Thus, the minus sign will be ignore an we have y ln(x + x 1). Therefore, cosh 1 x ln(x + x 1) since y cosh 1 x an the omain is the set of real number which is x 1. Derivatives of Hyperbolic Functions The ifferentiation formula for the hyperbolic functions. 1. x sinhx coshx. x coshx sinhx 3. x tanhx sech x 4. x cschx cschxcothx 5. x sechx sechxtanhx 6. x cothx csch x Example 4. Show that x sinhx coshx. Solution: x sinhx e x x (ex ) ex + e x cosh x Therefore, x sinhx coshx. Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

12 1 Derivatives an Its Application Example 4.3 Evaluate x cosh x. Solution: from erivative of hyperbolic function x cosh x sinh x x x sinh x x Derivatives of Inverse Hyperbolic Functions 1. x sinh 1 x 1 1+x. x cosh 1 x 1 x 1 3. x tanh 1 x 1 1 x 4. x csch 1 x 1 x 5. x sech 1 x 1 6. x coth 1 x 1 1 x x x +1 1 x Example 4.4 Prove that x sinh 1 x 1 1+x. Solution: Let y sinh 1 x x sinhy x y coshy, but y x 1 x y 1 coshy sinh y x Hence, x sinh 1 x 1 1+x. Example 4.5 Differentiate y sinh 1 x 3. Solution: Let h(x) sinh 1 x an g(x) x 3 But y x y h(g(x)) h (g(x))g (x) (g(x)) 3x 3x 1 + x 6 Hence, x sinh 1 x 3 3x 1+x Implicit Differentiation; Higher Orer Derivatives Implicit Differentiation The functions that we have met so far can be escribe by expressing one variable explicitly in terms of another variable. For example, y x or y sinx or, in general, y f (x). Some functions, however, Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

13 4.3 Implicit Differentiation; Higher Orer Derivatives 13 are efine implicitly by a relation between x an y such as x + y 5 or x 3 + y 3 6xy. This consists of ifferentiating both sies of the equation with respect to x an then solving the resulting equation for y. Example 4.6 If x + y 5, fin y x. Solution: Differentiate both sies of the equation x + y 5 x (x + y ) x 5 x x + x y 0 y x x y Example 4.7 Fin an equation of the line that is tangent to the circle x +y 9 at the point (, 5). Solution: The point (, 5) lies on the top half of the circle, which is the graph of a ifferential function y. From the above example, y x (, 5) 5 Therefore the slope of the tangent line we seek is 5, so an equation of the line is y 5 5 (x + ) 4.3. Higher Derivatives If f is a ifferentiable function, then its erivative f is also a function, so f may have a erivative of its own, enote by ( f ) f. This new function f is calle the secon erivative of f because it is the erivative of the erivative of f. Using Leibniz notation, we write the secon erivative of y f (x) as x (y x ) y x Another notation is f (x) D f (x). The thir erivative f is the erivative of the secon erivative: f ( f ). So f (x) can be interprete as the slope of the curve y f (x) or as the rate of change of f (x). The alternative notations for the thir erivative are y f (x) y x ( y x ) 3 y x 3 D3 f (x) The process can be continue. The fourth erivative f is usually enote by f (4). In general, the n th erivative of f is enote by f (n) an is obtaine from by ifferentiating n-times. If y f (x), we write y (n) f (n) (x) n y x n Dn f (x) Example 4.8 If f (x) xcosx, fin an interpret f (4) (x). Solution: Here f (x) xcosx, then we have f (x) xcosx f (x) cosx + xsinx f (x) sinx sinx xcosx sinx xcosx f (3) (x) cosx + cosx xsinx cosx xsinx f (4) (x) xcosx Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

14 14 Derivatives an Its Application Example 4.9 Fin f (55) (x), where f (x) cosx. Solution: Here f (x) cos x, then we have f (x) cosx f (x) sinx f (x) cosx f (3) (x) sinx f (4) (x) cosx The erivatives occur in a cycle of length 4 because of this f (55) (x) sinx. Example 4.30 Fin the n th erivative of f for n 1, where f (x) e x x. Solution: Using chain rule we have f (x) e x x f (x) e x x f (x) e x f (3) (x) e x f (4) (x) e x an f (n) (x) e x for n 3 Example 4.31 Fin the n th erivative of f, where f (x) e cx. Solution: By the chain rule, f (x) ce cx f (x) c e cx f (3) (x) c 3 e cx In general, for any positive integer n, f (n) (x) c n e cx 4.4 Applications of Derivative Tangent Line Approximation Definition If y f (x) is ifferentiable at x a, then L(x) f (a) + f (a)(x a) is the tangent line linear approximation of the curve t f (x). The approximation f (x) L(x) is the stanar linear approximation of f at a.the center of approximation is a. Example 4.3 Fin the linear approximation to the function f (x) 3 + x at x 1. Solution: From the above efinition use L(x) f (a) + f (a)(x a) for f (x) 3 + x. The erivative of f (x) is f (x) 1 3+x. Substitute a 1, then we have f (1) 1 4 an f (1). So, L(x) f (a) + f (a)(x a) + 1 (x 1) x Therefore, L(x) 1 4 x is the linear approximation of f (x) 3 + x near x 1. Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

15 4.4 Applications of Derivative 15 Example 4.33 Approximate 3.98 an 4. for a function f (x) 3 + x at x 1. Solution: For 3.98, f (0.98) an 0.98 is near a 1. So, f (1 + ( 0.0)) f (1) + f (1)(0.98 1) Therefore, the actual value of 3.98 is about For 4., f (1.) an 1. is near a 1. So, f (1.)) f (1) + f (1)(1. 1) Therefore, the actual value of 4. is about Maximum an Minimum Values Definition 4.4. A function f has an absolute maximum (or global maximum) at c if f (c) f (x) for all x in D, where D is the omain of f. The number f (c) is calle the maximum value of f on D. Similarly, a function f has an absolute minimum (or global minimum)at c if f (c) f (x) for all x in D an the number f (c) is calle the minimum value of f on D. The maximum an minimum values of f are calle the extreme values of f. f (c) is the maximum value of f if an only if (c, f (c)) is as high as any point on the graph of f an f (c) is the minimum value of f if an only if (c, f (c)) is as low as any point on the graph of f Definition A function f has a local maximum (or relative maximum) at c if f (c) f (x) when x is near c; that is, f (c) f (x) that for all x in some open interval containing c. Similarly, f has a local minimum at c if f (c) f (x) when x is near c. Example 4.34 The function f (x) cosx takes on its (local an absolute) maximum value of 1 infinitely many times, since cosnp 1 for any integer N an 1 cosx 1 for all x. Likewise, cos(n+1)p 1 is its minimum value, where n is any integer. Example 4.35 If f (x) x, then f (x) f (0) because x 0 for all x. Therefore, f (0) 0 is the absolute(an local)minimum value of f. This correspons to the fact that the origin is the lowest point on the parabola y x. However, there is no highest point on the parabola an so this function has no maximum value. A function f may or may not have extreme values on a set I, epening on f an on I. The following simple examples illustrate four of the possibilities 1. If f (x) x, then on [0,1] the function f has the maximum value on 1 an minimum value on 0. Figure 4.1: f (x) x on [0,1]. If f (x) tanx, then on ( π, π ) the function f has neither a maximum value nor a minimum value. 3. If f (x) x for 1 x < 0 an 0 < x 1, an if f (0) 1, then on [ 1,1] the function has the maximum value because f oes not assume the value If f (x) x for < x <, then on (, ) the function has neither a minimum value nor a maximum value. Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

16 16 Derivatives an Its Application Figure 4.: f (x) tanx Figure 4.3: f (x) x Figure 4.4: f (x) x, (, ) Theorem The Extreme Value Theorem If f is continuous on a close interval [a,b], then f attains an absolute maximum value f (c) an an absolute minimum value f () at some numbers c an in [a,b]. Theorem 4.4. Fermat s Theorem Suppose c is an interior point of an interval I, an f has a local maximum or minimum at c, an if f (c) exists, then f (c) 0. Definition A critical number of a function f is a number c in the omain of f such that either f (c) 0 or f (c) oes not exist. If f has a local maximum or minimum at c, then c is a critical number of f. Example 4.36 Fin the critical numbers of f (x) x 3 5 (4 x). Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

17 4.4 Applications of Derivative 17 Solution: The prouct rule gives f (x) 3 5 x 5 (4 x) + x 3 5 ( 1) 3(4 x) x 5 3 5x 5 3(4 x) 5x 5x 5 1 8x 5x 5 Therefore, f (x) 0 if 1 8x 0, that is x 3 an f (x) oes not exist when x 0. Thus, the critical numbers are 3 an 0. The Close Interval Metho To fin the absolute maximum an minimum values of a continuous function f on a close interval [a,b] Step 1: Fin the critical numbers of f in (a,b). Step : Fin the values of f at the en points of the interval an at critical points. Step 3: The largest of the values from Step is the absolute maximum value; the smallest of these values is the absolute minimum value Example 4.37 Fin the absolute maximum an absolute minimum of the function f (x) x + 4x + 5 on the interval [ 3,1]. Solution: The function f (x) x + +4x + 5 is continuous on the interval [ 3,1] Step 1: First we have to fin the critical points of f f (x) x x x 4 x is a critical value in ( 3,1) Step : Now we have to fin the values of f at the en points of the interval. f ( 3) ( 3) + 4( 3) + 5 f ( ) ( ) + 4( ) f (1) (1) + 4(1) Step 3: The largest of the values from Step is f (1) 10 is absolute maximum value; the smallest value f ( ) 1 is absolute minimum value. Example 4.38 Fin the absolute maximum an absolute minimum of the function f (x) x 3 + 3x 9x 7 on the interval [ 4,]. Solution: The function f (x) x 3 + 3x 9x 7 is continuous on the interval [ 4,] Step 1: First we have to fin the critical points of f f (x) 3x + 6x 9 0 3x + 6x 9 0 (x + 3)(x 1) 0 x 3 or x 1 are a critical value in ( 4,) Step : Now we have to fin the values of f at the critical point an at the en points of the interval. f ( 4) ( 4) 3 + 3( 4) 9( 4) 7 13 f ( 3) ( 3) 3 + 3( 3) 9( 3) 7 0 f (1) (1) 3 + 3(1) 9(1) 7 1 f () () 3 + 3() 9() 7 5 Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

18 18 Derivatives an Its Application Step 3: The largest of the values from Step is f ( 3) 0 is absolute maximum value; the smallest value f (1) 1 is absolute minimum value. The Mean Value Theorem To arrive at the Mean Value Theorem we first nee the following result. Theorem Rolle s Theorem Let f be a function that satisfies the following three hypotheses: 1. f is continuous on the close interval [a,b]. f is ifferentiable on the open interval (a,b) 3. f (a) f (b) Then there is a number c in (a,b) such that f (c) 0. Example 4.39 Prove that the equation x 3 + x 1 0 has exactly one real root. Solution: First we use the Intermeiate Value Theorem to show that a root exists. Let f (x) x 3 + x 1. Then f (0) 1 < 0 an f (1) 1 > 0. Since f is a polynomial, it is continuous, so the Intermeiate Value Theorem states that there is a number c between 0 an 1 such that f (c) 0. Thus, the given equation has a root. To show that the equation has no other real root, we use Rolle s Theorem an argue by contraiction. Suppose that it ha two roots a an b. Then f (a) 0 f (b) an, since f is a polynomial, it is ifferentiable on (a,b) an continuous on [a,b]. Thus, by Rolle s Theorem, there is a number c between a an b such that f (c) 0. But f (x) 3x for all x (since x 0 ) so f (x) can never be 0. This gives a contraiction. Therefore, the equation can not have two real roots. Theorem The Mean Value Theorem Let f be a function that satisfies the following hypotheses: 1. f is continuous on the close interval [a,b]. f is ifferentiable on the open interval (a,b) Then there is a number c in (a,b) such that f (c) f (b) f (a) or, equivalently f (b) f (a) f (c)(b a) b a Example 4.40 Let f (x) 1 3 x3 + x. Fin a number c in (0,3) such that f f (3) f (0) (c) 3 0 Solution: Since f (3) f (0) We seek a number c in (0,3) such that f (c) 5. But f (x) x + So that c must satsfy c + 5. Therefore, c 3 an since c must be in (0,3), we conclue that c 3. Example 4.41 Show that e x 1 + x for x 0. Solution: Let f (x) e x, since f is continuous an ifferentiable for x 0. By mean value theorem there is c in (0,x) such that f (c) ex e 0 x 0 ec ex 1 x clearly, e c 1 for c 0, then e x 1 x 1 e x 1 x e x 1 + x Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

19 4.4 Applications of Derivative 19 Example 4.4 Show that sina sinb a b for all a an b. Solution: Let f (x) sin x, since f is continuous an ifferentiable for all x. Let a > b. By mean value theorem there is c in (a,b) such that Since, 1 cosc 1, then f (c) f (a) f (b) a b cosc sina sinb a b sina sinb 1 1 a b (a b) sina sinb (a b) sina sinb a b Similarly, for a < b we have sina sinb a b. Example 4.43 Show that b a b ln( b a ) b a a for 0 < a < b. Solution: Let f (x) lnx, since f is continuous on [a,b] an ifferentiable on (a,b). By mean value theorem there is c in (a,b) such that Since, 0 < a < c < b, then f (c) f (b) f (a) b a 1 a > 1 c > 1 b 1 a b a a b a b 1 c > lnb lna b a lnb lna b a > 1 b > lnb lna > b a b ln( b a ) b a a Example 4.44 Suppose that we know that f (x) is continuous an ifferentiable on [6,15]. Let s also suppose that f (6) an f (x) 10. What is the largest possible value for f (15)? Solution: Let s start with the conclusion of the mean value theorem f (c) f (15) f (6) 15 6 f (15) f (6) f (c)(15 6) substituting the known values an rewriting this gives f (15) f (6) f (c)(15 6) + 9 f (c) Now, we know that f (x) 10. So in particular we know that f (c) 10. This gives us the following f (15) + 9 f (c) + (9)(10) 88 All we i was replace f (c) with its largest possible value. This means that the largest possible value for f (15) is 88. Increasing an Decreasing Test Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

20 0 Derivatives an Its Application Theorem Let f be continuous on an interval I. If f (x) exists an equals 0 for each interior point x of I, then f is constant on I.. Let f an g be continuous on an interval I. If f (x) an g (x) exist an are equal for each interier point x of I, then f g is constant on I. In otherwors, there is a constant c such that f (x) g(x) + c for all I. Theorem Let f be continuous on an interval I an ifferentiable at each interier point of I. 1. If f (x) > 0 at each interior point of I, then f is increasing on I. Moreover, f is increasing on I if f (x) > 0 except for a finite number of points x in I.. If f (x) < 0 at each interior point of I, then f is ecreasing on I. Moreover, f is ecreasing on I if f (x) < 0 except for a finite number of points x in I. Example 4.45 Fin the interval where the function f (x) x is increasing an the interval where it is ecreasing. Solution: The erivative of f (x) x is f (x) x > 0 when x > 0, an f (x) x < 0 when x < 0, f is increasing in the interval (0, ) an ecreasing in the interval (,0). Example 4.46 Determine the interval where the function f (x) x 3 3x 4x + 3 is increasing an where it is ecreasing. Solution: The erivative of f is f (x) 3x 6x 4 3(x + )(x 4) an it is continuous everywhere. The zeros of f (x) are x an x 4. So, f (x) > 0 in the interval (, ), f (x) < 0 in the interval (,4) an f (x) > 0 in the interval (4, ). Therefore, f is increasing on the intervals (, ) an (4, ) an it is ecreasing on the interval (,4). The First Derivative Test Theorem Let f be ifferentiable on an open interval about the number c except possibly at c, where f is continuous. Suppose that c is a critical number of a continuous function f. 1. If f changes from positive to negative at c, then f has a local maximum at c.. If f changes from negative to positive at c, then f has a local minimum at c. 3. If f oes not change sign at c (for example, if f is positive on both sies of c or negative on both sies), then f has no local maximum or minimum at c. Concavity Test Theorem Assume that f is exist on an open interval I. 1. If f (x) > 0 for all x in the interval I, then the graph of f is concave upwar on I.. If f (x) < 0 for all x in the interval I, then the graph of f is concave ownwar on I. Definition A point P on a curve y f (x) is calle an inflection point if f is continuous there an the curve changes from concave upwar to concave ownwar or from concave ownwar to concave upwar at P. The Secon Derivative Test Theorem Suppose f is continuous near c. 1. If f (c) 0 an f (c) > 0, then f has a local minimum at c.. If f (c) 0 an f (c) < 0, then f has a local maximum at c. 3. If f (c) 0, then from this test alone, we can not raw any conclusion about a relative extreme value of f at c. Example 4.47 Determine the relative extreme of the function f (x) x 3 3x 4x + 3 using the secon erivative test. Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

21 4.4 Applications of Derivative 1 Solution: We have f (x) 3x 6x 4 3(x + )(x 4) So, f (x) 0 gives x an x 4, the critical point of f. Next we compute f (x) 6x 6 6(x 1) Since f ( ) 6( 1) 18 < 0. The secon-erivative test implies that the point (,60) as a relative maximum of f. Also f (4) 6(4 1) 18 > 0 an the secon erivative test implies that the point (4, 48) is a relative minimum of f, which conforms the result obtaine. Example 4.48 Fin the absolute maximum value of the function f (x) 1 x 9 x on the interval (0, ). Solution: First we have to fin the critical values on the interval f (x) x 0 9 x 1 x 9 x ±3 The only critical value in the interval (0, ) is x 3. Next fin the secon erivative of f (x) f (x) ( 9 x 3 ) 18 x 3 f (3) < 0 Therefore, f (3) is the absolute maximum value of f on (0, ). Example 4.49 Fin the absolute maximum value of the function f (x) 4 x 8 x, where x > 0. Solution: First we have to fin the critical values by first erivative f (x) solving for x we get x 8 x x 4 x ± Therefore, x is on the interval x > 0. Next check whether the secon orer conition is satisfie or not. f (x) ( 8 x 3 ) 16 x 3 f () < 0 Therefore, f () 4 () 8 16 is the absolute maximum value of f (x) 4 x 8 x on the interval x > 0. Example 4.50 Let f (x) x 3 3x. Using the secon erivative test, fin the relative extreme values of f. Solution: By ifferentiation we obtain f (x) 3x 3 3(x 1)(x + 1) an f (x) 6x Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

22 Derivatives an Its Application Therefore f (x) 0 when x 1 or x 1. Since f ( 1) 6 < 0 an f (1) 6 > 0 We know from the secon erivative test that f ( 1) 0 is a relative maximum values of f, whereas f (1) 4 is a relative minimum values of f. These are the only relative extreme values of f. 4.5 Ineterminate Forms an L Hopital s Rule Ineterminate Forms Ineterminate Quotient L Hopital Rule: Suppose f an g are ifferentiable an g (x) 0 near a (except possibly at a). Suppose that x a x a x a f (x) 0 an g(x) 0 or x a f (x) ± an x a f (x) g(x) 0 or 0 x a g(x) ± ; that is f (x) g(x) Then f (x) x a g(x) f (x) x a g (x) if the it on the right sie exists (or is or ). L Hopital s Rule is also vali for one-sie its an for its at infinity or negative infinity; that is, x a can be replace by any of the symbols x a +, x a, x or x. Example 4.51 Solve x 1 x 0 x. Solution: Since x 0 ( x 1) 0 an x 0 x 0 By L Hopital rule x 1 0 x 0 x 0 is 0 0 form x 1 x 0 x x (x 1) x 0 x x x ln x 0 1 ln x x 0 ln Therefore, x 0 x 1 x ln. Example 4.5 Solve lnx x 1 x 1. Solution: Since x 1 lnx 0 an x 0 (x 1) 0 lnx x 1 x is 0 0 form Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

23 4.5 Ineterminate Forms an L Hopital s Rule 3 By L Hopital rule lnx x 1 x 1 x 1 x x lnx (x 1) 1 x x 1 1 x 1 x 1 1 Therefore, x 1 lnx x 1 1. e Example 4.53 Solve x. x x Solution: Since e x an x x x By L Hopital rule e x x x is form x e x x x x ex x x e x x 1 x is form Again we apply L Hopital rule e x x x x x 1 e x x ex x x e Therefore, x. x 1 x Ineterminate Proucts If f (x) 0 an g(x) ±, then it isn t clear what the value of f (x)g(x) is calle an ineterminate form of type 0 x a x a x a. Example 4.54 Evaluate xlnx. x 0 + Solution: Since x 0 an lnx x 0 + x 0 + xlnx x 0 + x 0 + x 0 + lnx is x x type x x 1 x x 0 + x 0 by L Hopital Rule Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

24 4 Derivatives an Its Application Ineterminate Differences If f (x) an g(x), then the it ( f (x) g(x)) is calle is calle an ineterminate x a x a x a form of type. Example 4.55 Evaluate tanx). x ( π ) (secx Solution: Since secx an tanx ) ) x ( π x ( π tanx) secx tanx type ) (secx x ( π ) x ( π ) x ( π 1 x ( π ) ( cosx sinx cosx ) is type 1 sinx is 0 x ( π ) cosx 0 type cosx by L Hopital Rule x ( π ) sinx 0 Ineterminate Powers Several ineterminate forms arise from the it x a ( f (x)) g(x) 1. f (x) 0 an g(x) 0 type 0 0 x a x a. f (x) an g(x) 0 type 0 x a x a 3. f (x) 1 an g(x) ± type 1 ± x a x a Each of these three cases can be treate either by taking the natural logarithm Let y f (x) g(x), then lny g(x)ln f (x) or by writing the function as an exponential f (x) g(x) e g(x) ln f (x). Example 4.56 Calculate x 0 +(1 + sin4x)cotx. Solution: Since + sin4x) 1 an x 0 +(1 cotx x 0 + x 0 +(1 + sin4x)cotx + sin4x) cotx x 0 x 0 +(1 + is 1 type e x 0 + ln(1+sin4x)cotx e x 0 + cotxln(1+sin4x) e ln(1+sin4x) x 0 + tanx is 0 0 type e x 0 + e 4 4cos4x 1+sin4x sec x by L Hospital Rule Example 4.57 Evaluate x 0 + xx. Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

25 4.5 Ineterminate Forms an L Hopital s Rule 5 Solution: Notice that this is ineterminate form of 0 0. So by ineterminate form of power, we have x 0 + xx x 0 + elnxx x 0 + exlnx e x 0 + xlnx e 0, since x 0 + xlnx 0 1 Curve Sketching Guielines for Sketching a Curve The following checklist is intene as a guie to sketching a curve y f (x) by han. Not every item is relevant to every function. For instance, a given curve might not have an asymptote or possess symmetry. But the guielines provie all the information that nee to make a sketch that isplays the most important aspects of the function. 1. Domain It s often useful to start by etermining the omain D of f, that is, the set of values of x for which f (x) is efine.. Intercepts Fin x an y intercept. 3. Symmetry (a) If f ( x) f (x) for all x in D, the function is an even function an the curve is symmetric about the y-axis. (b) If f ( x) f (x) for all x in D, then f is an o function an the curve is symmetric about the origin. (c) If f (x + p) f (x) for all x in D, where p is a positive constant, then f is calle a perioic function an the smallest such number is calle the perio. For instance, y sinx has perio p. 4. Asymptotes (a) Horizontal Asymptotes. If either f (x) L or f (x) L, then the line y L is a horizontal asymptote of the x x curve y f (x). If it turns out that f (x) ±, then we o not have an asymptote to the x right. (b) Vertical Asymptotes. The line x a is a vertical asymptote if at least one of f (x), f (x), x a x a + f (x) or f (x) is true. For rational functions you can locate the vertical x a x a + asymptotes by equating the enominator to 0 after canceling any common factors. But for other functions this metho oes not apply. (c) Oblique Asymptotes. Some curves have asymptotes that are oblique, that is, neither horizontal nor vertical. If ( f (x) (mx + b)) 0, then the line y mx + b is calle an oblique/slant asymptote. For x rational functions, slant asymptotes occur when the egree of the numerator is one more than the egree of the enominator. In such a case the equation of the oblique asymptote can be foun by long ivision. 5. Intervals of Increase or Decrease Compute f (x) an fin the intervals on which f (x) is positive (f is increasing) an the intervals on which f (x) is negative (f is ecreasing). 6. Local Maximum an Minimum Values 7. Concavity an Points of Inflection Compute f (x) an use the Concavity Test. Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

26 6 Derivatives an Its Application 8. Sketch the Curve Using the information in items 1 7, raw the graph. Sketch the asymptotes as ashe lines. Example 4.58 Use the guielines to sketch the curve y x x 1. Solution: 1. The omain is {x x 1 0} {x x ±1} (, 1) ( 1,1) (1, ). The x-an y-intercept are both Since f ( x) f (x), the function f is even. The curve is symmetric about the y-axis. 4. The line y is a horizontal asymptote since 5. x ± x x 1 x ± 1 1 x since the enominator is 0 when x ±1, the it of y as x approaches to 1 + an 1 is, respectively an the it of y as x approaches to 1 an 1 + is, respectively. Therefore the lines x 1 an x 1 are vertical asymptotes. f (x) 4x(x 1) x x (x 1) 4x (x 1) Since f (x) > 0 when x < 0 (x 1) an f (x) < 0 when x > 0 (x 1), f is increasing on (, 1) an ( 1,0) an ecreasing on (0,1) an (1, ). 6. The only critical number is x 0. Since f changes from positive to negative at 0, f (0) is a local maximum by first erivative test. 7. f (x) 4(x 1) + x (x 1) (x 1) 4 1x + 4 (x 1) 3 Since 1x + 4 > 0 for all x, we have f (x) > 0 x 1 > 0 x > 1 an f (x) < 0 x < 1. Thus the curve is concave upwar on the intervals (, 1) an (1, ) an concave ownwar on ( 1,1). It has no point of inflection since 1 an -1 are not in the omain of f. 8. Sketch the Curve Using the information in items 1 7, raw the graph. Sketch the asymptotes as ashe lines. Figure 4.5: f (x) x x 1 Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

27 4.5 Ineterminate Forms an L Hopital s Rule 7 Application Problems 1. Rate of Change The first interpretation/application of a erivative is rate of change. If f (x) represents a quantity at any x then the erivative f (a) represents the instantaneous rate of change of f (x) at x a. Example 4.59 Suppose that the amount of water in holing tank at t minutes is given by V (t) t 16t + 0. Determine each of the following (a) Whether the volume of water in the tank increasing or ecreasing at t minute? (b) Whether the volume of water in the tank increasing or ecreasing at t 6 minute? (c) Is the volume of the water in the tank changing faster at t or t 6 minute? () Is the volume of the water in the tank ever change? If so, when? Solution: First compute the rate of change of the volume by fining the erivative of this function since that give a formula for the rate of change at any time t. So, V t 4t 16 By first erivative test if the rate of change is positive, then the quantity will be increasing an if the rate of change is negative, then the quantity will be ecreasing. (a) at t minute, the rate of change of the volume V () 8. So, at t the rate of change is negative an the volume must be ecrease at this time. (b) at t 6 minute, the rate of change of the volume V (6) 8. So, at t 6 the rate of change is positive an the volume must be increase at this time. (c) To answer this question all that we look at is the size of the rate of change an we o not worry about the sign of the rate of change. All that we nee to know here is the larger the number the faster the rate of change. () The volume will not be changing if it has a rate of change of zero. In orer to have a rate of change of zero this means that the erivative must be zero. So, to answer this question we will then nee to solve V 4t 16 0 t 4 t So, at t 4 the volume is not changing.. Velocity Definition If f (t) is the position of a particle moving on a coorinate line, then the instantaneous velocity of the particle at time t is efine by v(t) v (t) v t. Example 4.60 Suppose that the position of an object after t hours is given by, f (t) t t + 1 (a) Is the object moving to the right or the left at t 9? (b) Does the object ever stop moving? Solution: The erivative is f (t) 1. (t+1) (a) To etermine if the object is moving to the right (Velocity is positive) or left (Velocity is negative) we nee the erivative at t 9. f (9) 1 (10) So, the velocity at t 9 is positive an so the object is moving to the right at t 9. (b) The object will stop moving if the velocity is ever zero. However, note that the only way a rational expression will ever be zero is if the numerator is zero. Since the numerator of the erivative an hence the spee is a constant it can not be zero. Therefore, the velocity will never stop moving. In fact, the object will always be moving to the right since velocity is always positive. 3. Acceleration Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

28 8 Derivatives an Its Application Definition 4.5. If f (t) is the position function of a particle moving on a coorinate line, then the instantaneous acceleration of the particle at time t is efine by a(t) v (t) v t or alternatively, since v(t) f (t), a(t) f (t) f t Example 4.61 Let f (t) t 3 6t be the position function of a particle moving along a y-axis, where f is in meters an t is in secons. Fin the instantaneous acceleration a(t). Solution: a(t) v t 6t 1. Remark Interpreting the sign of acceleration.a particle in rectilinear motion is speeing up when its velocity an acceleration have the same sign an slowing own when they have opposite signs. 4. Optimization Problem Example 4.6 An open-top box is to be mae by cutting small congruent squares from the corners of a 1 by 1 meters sheet of tin an bening up the sies. How large shoul the squares cut from the corners be to make the box hol as much as possible? Solution: In the figure below, the corner squares are x meter on a sie. The volume of the box is a function of this variable v lwh v(x) x(1 x)(1 x) for 0 x 6 v x x + 1x v 0 if an only if x or x 6 x Thus, v() 18, v(0) v(6) 0. Therefore, v() 18m 3 is the maximum volume obtaine by Figure 4.6: 1 m by 1 m square sheet cut out squares m on each sie. Example 4.63 Four meter of wire is to be use to form a square an a circle. How much of the wire shoul be use an how much shoul be use for the circle to enclose the maximum total area? Solution: The total area is given by let x be the sie of the square an r are raius of the circle A Area of Square + Area of circle x + πr (4.1) an the sum perimeter of square an circumference of circle is 4, Since the total amount of wire is 4 m, 4x + πr 4, thus r (1 x) π for 0 x 1 (4.) Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

29 4.5 Ineterminate Forms an L Hopital s Rule 9 By substituting equation(4.) into equation (4.1), we have A(x) x + π( A x (1 x) ) π 1 ((π + 4)x 8) Thus, the only critical number is x π , since A(0) 1.73, A(0.56) 0.56 an A(1) 1. Therefore, the maximum area occurs when x 0; that is, all the wire is use for the circle. Example 4.64 You have been aske to esign a 1 liter(1000cm 3 ) oil-can shape like a right circular cyliner. What imensions will use the least material? Solution: Let the right circular cyliner can have height h an iameter r as shown in the figure below. Here h an r are relate by the equation an total surface area is given by πr h 1000 (4.3) Figure 4.7: Cyliner having raius r an height h A πr + πrh (4.4) solving equation (4.3) for h, we have Thus, h 1000 πr (4.5) A πr + πrh πr + πr( 1000 πr ) πr r A(r) πr r A r 4πr 000 r 0 4πr 000 4πr r π r by setting A r 0 Therefore a critical point is r π. If the omain of A were a close interval, we coul fin out by evaluating A at this critical point an the enpoints an comparing the results. But the omain is Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

30 30 Derivatives an Its Application not a close interval. So, we have to check at r π A 4πr 000 r r A r 4π r 3 by secon erivative test. The secon erivative is positive throughout the omain of A. The value of a at r π therefore an absolute minimum because the graph of A is concave up an is h 1000 πr π( π ) π r (4.6) The above equation (4.6) tells us that the most efficient can has its height equal to its iameter; that is, r 5.4cm an h 10.84cm. Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

31 4.6 Exercise Exercise 1. Fin the erivative of f (x) 5x 9x using the efinition of erivatives. State the omain of the function an the omain of its erivative. Ans. f (x) 5 18x, R, R. Let l be the tangent line to the parabola y x at the point (1,1). The angle of inclination of l is the angle φ that l makes with the positive irection of the x-axis. Calculate φ correct to the nearest egree. 3. Fin the erivative of y (x + 1)(x 3 + 1). 4. Differentiate y x 1+ x. Ans. 63 o Ans. 5x 4 + 3x + x Ans. y x(4+ x) (1+ x) 5. Fin an equation of the tangent line to the curve y x x+1 at the point (1,1). Ans. y 1 x Fin an equation of the tangent line an normal line to the curve y 3x+1 at the point (1,). x +1 Ans. y 1 x + 5, y x 7. Fin the first an secon erivative of the function f (x) 1+x x. xx Ans. y, f (x) (1+x) (1+x) 3 8. If f (x) xg(x), where g(4) 8 an g (4) 7, fin f (4). Ans Show that the curve y 6x 3 + 5x 3 has no tangent line with slope Fin an equation of the normal line to the parabola y x 5x + 4 that is parallel to the line x 3y 5. Ans. y 1 3 x For what value of x is the function f (x) x 9 ifferentiable? fin a formula for f. { x if x > 3 Ans. No ifferentiation at x 3 an x 3, f (x) x if x < 3 1. For what value of a an b is the line x + y b tangent to the parabola y ax when x? 13. Fin the value of c such that the line y 3 x + 6 is tangent to the curve y c x. 14. Differentiate f (θ) secθ 1+secθ. Ans. a 1, b Ans. f (θ) secθ tanθ (1+secθ) 15. Fin the it x 0 sin3x x. Ans If f (x) g(h(x)), where g( ) 8, g ( ) 4, g (5) 3, h(5), an h (5) 6, fin f (5). Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et

32 3 Derivatives an Its Application Ans Fin the implicit ifferentiation y x for a function xy 1 + x y. Ans. y 4xy xy y x x xy 18. If f (x) + x [ f (x)] 3 10 an f (1), fin f (1). Ans Fin the local maximum an minimum values of f (x) x 5 5x + 3 using both the first an secon erivative test 0. Let f (x) x, then fin x 1 (a) omain (b) intercept (c) symmetry () all the existing asymptotes (e) interval of increase an ecrease (f) local maximum an minimum values (g) interval of concavity an inflection point (h) sketch the graph of f Ans. Local maximum f ( 1) 7 an local maximum f (1) 1 Figure 4.8: f (x) x x 1 Any one can get this soft-copy from Google site Exous4Wisom natnael.nigussie@aastu.eu.et