Math Chapter 2 Essentials of Calculus by James Stewart Prepared by Jason Gaddis

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1 Math Chapter 2 Essentials of Calculus by James Stewart Prepare by Jason Gais Chapter 2 - Derivatives 21 - Derivatives an Rates of Change Definition A tangent to a curve is a line that intersects the curve at precisely one point Example Fin an equation of the tangent line to the parabola y = x 3 at the point P = ( 1, 1) Definition The tangent line to the curve y = f(x) at the point P (a, f(a)) is the line through P with slope provie that this limit exists f(x) f(a) m = lim, x a x a Note Alternatively, we can use the following limit: f(a + h) f(a) m = lim Example Fin an equation of the tangent line to the curve y = 2x (x+1) 2 at the point (0, 0) Note We can use the tangent problem to moel average velocity If we start at time a an go until time a + h, then change in the function is f(a + h) f(a) Hence, average velocity = isplacement time = f(a + h) f(a) h Definition We efine velocity (or instantaneous velocity) v(a) at time t = a to be the limit of the average velocities at t = a This is, f(a + h) f(a) v(a) = lim Example The isplacement (in meters) of a particle moving in a straight line is given by s = t 2 8t + 18, where t is measure in secons Fin the average velocity over each time interval: Fin the instantaneous velocity when t = 4 [3, 4]; [35, 4]; [4, 5]; [4, 45] Definition The erivative of a function f at a number a, enote by f (a) is provie this limit exists Note This can be rewritten as f f(a + h) f(a) (a) = lim, f f(x) f(a) (a) = lim x a x a Definition Fin the erivative of the function f(x) = 1 x 3 at the number a = 0 Note We can reefine the tangent line as follows: The tangent line to 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 1

2 Example Fin an equation of the tangent line to the parabola y = 1 x 3 at the point (0, 1) Definition Let y = f(x) We efine the ifference quotient to be y x = f(x 2) f(x 1 ) x 2 x 1 This is also known as the average rate of change of y with respect to x over the interval [x 1, x 2 ] Definition We efine the instantaneous rate of change of y with respect to x at x = x 1 as y lim x 0 x = lim f(x 2 ) f(x 1 ) x 2 x 1 x 2 x 1 Note The erivative f (a) is the instantaneous rate of change of y = f(x) with respect to x when x = a 2

3 22 - The erivative of a function Note We will now switch from thinking about the erivative at a point to the erivative of the entire function If we let x vary, then we say the erivative of f is, f (x) = lim h 0 f(x + h) f(x) h Example (pg 91, example 2) Example If f(x) = x + x, fin a formula for f (x) State the omain of f Note We will sometimes express the notation for f in ifferent ways: f (x) = y = y x = f x = x f(x) = Df(x) = D xf(x) The symbols D an /(x) are calle ifferentiation operators Definition A function f is ifferentiable at a if f (a) exists It is ifferentiable on an open interval (a, b) if it is ifferentiable at every number in the interval (Note: We can have a = an b = ) Example (pg 87, example 5) Where is the function f(x) = x ifferentiable Theorem If f is ifferentiable at a, then f is continuous at a Note The converse of the previous theorem is false See the previous example Note As a result of this theorem, we now know that if f is not continuous at a point, then it is not ifferentiable However, a function can be continuous an still fail to be ifferentiable If a curve has a vertical tangent line when x = a, then the curve is not ifferentiable at a That is, f is continuous at a an lim f (x) = x a Definition A secon erivative of f is the erivative of the first erivative Similarly, a thir erivative of f is the erivative of the secon erivative Example (pg 90, example 6) If f(x) = x x, fin an interpret f (x), f (x), anf (4) (x) 3

4 23 - Basic Differentiation Formulas Note In this section, we will etermine several shortcuts for fining erivatives It is important to always remember the limit efinition of a erivative, as not all erivatives can be foun this way Prop Differentiation formulas Let c be a constant an f, g ifferentiable functions, then we have the following rules: Derivative of a constant function: The power rule: x (xn ) = nx n 1 The constant multiple rule: The sum/ifference rule: Example Fin the erivatives of the following funtions: (a) f(x) = 7x 9 (b) f(x) = 1 x 3 (c) f(x) = 5 x 3 () f(x) = x 7 3x 5 + 2x x (c) = 0 where n is any real number x [cf(x)]c x f(x) [f(x) ± g(x)] = x x f(x) ± x g(x) Definition The normal line to a curve C at a point P is the line through P perpenicular to the tangent line at P Example Fin the equations of the tangent line an normal line to the curve y = (1 + 2x) 2 at the point (1, 9) Note Recall that horizontal tangent lines occur when the erivative is zero Example Fin the points on the curve y = x 3 + 3x 2 + x + 3 where the tangent line is horizontal Note Recall that we have the following limits from section 14: sin θ cos θ 1 lim = 1 lim = 0 θ 0 θ θ 0 θ Prop Derivatives of trig functions sin(x) = cos(x) x cos(x) = sin(x) x Example Differentiate y = 2 cos θ 3 sin θ Example The equation of motion of a particle is s = 2t 3 7t 2 + 4t + 1, where s is in meters an t is in secons Fin the velocity an acceleration as functions of t Fin the acceleration after 1 secon 4

5 24 - Prouct an Quotient Rule Prop The Prouct Rule If f an g are both ifferentiable, then [f(x)g(x)] = f(x) [g(x)] = g(x) x x x [f(x)] Example Differentiate 3 x cos(x) Prop The Quotient Rule If f an g are both ifferentiable, then x [ ] f(x) = g(x) x [f(x)] f(x) x [g(x)] g(x) [g(x)] 2 Example Differentiate sin(x) x Example Differentiate x3 +2x 1 x 2 +2 Prop Derivatives of Trig Functions (sin x) = cos x (cos x) = sin x x x x (tan x) = sec2 x x (cot x) = csc2 c (csc x) = csc x cot x (sec x) = sec x tan x x x Example Differentiate f(x) = 1 sec x tan x For what values oes the graph of f have a horizontal tangent? 5

6 25 - The Chain Rule Prop The Chain Rule If f an g are both ifferentiable an F = f g = f(g(x)), then F is ifferentiable an F is given by the prouct F (x) = f (g(x)) g (x) In other (Leibniz) notation, if u = g(x) an y = f(u), then y x = y u u x Example Let F (x) = x 3 + 3x + 2 Fin F (x) Example Let f(x) = x 2 an g(x) = sin x Fin (f g) (x) an (g f) (x) Prop The Power Rule Combine with the Chain Rule If n is any real number an u = g(x) is ifferentiable, then x (un n 1 u ) = nu x Alternatively, x [g(x)]n = n[g(x)] n 1 g (x) Example Differentiate y = (x 2 + 1) 3 (2x 3 + 3x 4) 5 ( 7 x+1 Example Differentiate f(x) = 3x 2) (Remember: tan 2 x sec 2 x = 1) 6

7 26 - Implicit Differentiation Note Consier the equation of a circle of raius 4: x 2 +y 2 = 100 If we wante to ifferentiate this equation we woul first have to solve for y an evaluate the positive an negative square root There is an alternate metho calle implicit ifferentiation Remark To use implicit ifferentiation, we ifferentiate both sies with respect to x an solve for y Remember that y is a funtion with respect to x Example Fin y if x 2 + y 2 = 100 Fin an equation of the tangent line at (6, 8) Example Fin y if y 5 + x 2 y 3 = 1 + x 4 y Example Fin y if y sin(x 2 ) = x sin(y 2 ) Example Use implicit ifferentiation to fin an equation of the tangent line to the curve y = x 2 + 2xy y 2 + x = 2 at the point (1, 2) 7

8 Sie Note Note Two erivatives that are very important but not mentione in chapter 2 are those of e x an ln x, the exponential function an the natural log function, respectively You have seen these functions before in an algebra class an we will spen more time with them in chapter 5, but they are worth mentioning now Remark Recall that e = 2718 an ln 1 = 0 Also, these two functions are inverses of each other, so e ln x = x an ln(e x ) = x The omain of e x is R an its range is (0, ) The omain of ln x is (0, ) an its range is R Note We will first try to calculate x ex Remember that for any number a we have a m+n = a m a n e x+h e x x ex = lim e x e h e x = lim = lim e x eh 1 = e x e h 1 lim because e x oes not epen on the limiting variable, h The question now becomes: What is that limit? In one of Stewart s other books, Early Transcenentals, it is a efinition Definition The function e x is the unique function such that e h 1 lim = 1 Note If we use this as the efinition, then we get that x ex = e x e h 1 lim = e x 1 = e x Hence, the function e x is a function whose erivative is itself The only other function (that I am aware of) where this property hols is the constant zero function Example Let y = e x2 Fin y We will use the chain rule with f(x) = e x an g(x) = x 2 Then f (x) = e x an g (x) = 2x Hence, y = f (g(x)) g (x) = e x2 2x = 2xe x2 Note Now we turn to the next function, ln x We will use the property that e ln x = x an take the erivative of each sie See if you can spot where we will cheat in this proof x eln x = x x e ln x ln x = 1 by the chain rule x x ln x = 1 e ln x = 1 x The chain rule requires that both the insie an outsie functions are ifferentiable, an we on t know that ln x is ifferentiable As it turns out, it is (Theorem 517) an so this proof is fine Example Let y = ln x 2 Fin y Again, we will use the chain rule with f(x) = ln x an g(x) = x 2, so f (x) = 1 x an g (x) = 2x Then y = f (g(x)) g (x) = 1 x 2 2x = 2 x 8

9 Practice Problems 1 Differentiate each function (a) f(x) = 5e x + 3 (b) g(x) = e x (c) y = e x () h(x) = ex x+e x (e) f(t) = sin(e t ) 2 Fin an equation of the tangent line an normal line to the curve y = 2xe x at the point (0, 0) 3 Differentiate each function (a) f(x) = 3 ln x + 4x (b) g(x) = cos(ln x) (c) f(t) = 1+ln t 1 ln t () h(x) = ln x 1+x (e) y = ln(e x + xe x ) 4 Fin an equation of the tangent line to the curve y = ln(ln x) at the point (e, 0) 9

10 27 - Relate Rates Note Sometimes is becomes necessary to take erivatives with respect to a thir variable This occurs often in problems with relate rates Example Each sie of a square is increasing at a rate of 6cm/s At what rate is the area of the square increasing when the area of the square is 16cm 2? Example A spotlight on the groun shines on a wall 12 m away If a man 2 m tall walks from the spotlight towar the builing at a spee of 16 m/s, how fast is the length of his shaow on the builing ecreasing when he is 4 m from the builing? Example Water is leaking out of an inverte conical tank at a rate of 10,000 cm 3 /min at the same time that water is being pumpe into the tank at a constant rate The tank has hieght 6m an the iameter at the top is 4m If the water level is rising at a rate of 20 cm/min when the height of the water is 2m, fin the rate at which water is being pumpe into the tank If C is the rate the water is pumpe in, then V t = C 10, 000, where V = 1 3 πr2 h is the volume at time t By similar triangles, When h = 2 m, h t r 2 = h 6 r = 1 3 V = 1 ( ) π 3 h h = π 27 h3 V t = π h h2 9 t = 20 cm/min, so C 10, 000 = π 9 (200)2 (20) C = 10, , 000 π 289, 253cm 3 /min 9 10

11 28 - Linear Approximations an Differentials Note We can use tangent lines to approximate a curve at a specific point If we allow for a small amount of error, we can approximate the curve on some interval (though, that error grows as we make the interval larger) This is calle linear approximation Definition The linear approximation (or tangent line approximation) of f at a is f(x) f(a) + f (a)(x a) The linearization of f at a is the linear function whose graph is the tangent line, that is, L(x) = f(a) + f (a)(x a) Example Fin the linearization of the function f(x) = sin x at a = π 6 a, L(x) = 0591, the actual value is 0588) use it to approximate sin( π 5 ) (At Example Use linearization to approximate 998 Definition If y = f(x), where f is a ifferentiable function, then the ifferential x is an inepenent variable The ifferential y is then efine in terms of x by the equation y = f (x)x Example Fin the ifferential of y = x 1+x 2 Note The efintion means that we can give x the value of any real number (See graph on pg 136) Definition The maximum error is given by y The relative error is given by y y by y y The percentage error is the relative error expresse as a percent an can be approximate Example The raius of a circular isk is given as 24 cm with a maximum error in measurement of 02 cm Use ifferentials to estimate the maximum error in the calculate area of the isk What is the relative error? What is the percentage error? (relative error) A A A = πr 2 A = 2πrr = 2π(24)(02) = 96π 30cm 2 (max error) A A = 2πrr πr 2 = 2r πr = 2r = 2(02) = 1 r = 0016cm2 (percentage error) = = 16% 11