Chapter 5 Logarithmic, Exponential, and Other Transcendental Functions

Transcription

1 Chapter 5 Logarithmic, Exponential, an Other Transcenental Functions 5.1 The Natural Logarithmic Function: Differentiation 5.2 The Natural Logarithmic Function: Integration 5.3 Inverse Functions 5.4 Exponential Functions: Differentiation an Integration 5.5 Bases Other than e an Applications 5.6 Ineterminate Forms an L Hopital s Rule 5.7 Inverse Trigonometric Functions: Differentiation 5.8 Inverse Trigonometric Functions: Integration 5.9 Hyperbolic Functions

2 5.7. Inverse Trigonometric Functions: Differentiation Definitions of Inverse Trigonometric Functions Review Function Domain Range y = arcsin x = sin 1 x iff sin y = x 1 x 1 π 2 y π 2 y = arccos x = cos 1 x iff cos y = x 1 x 1 0 y π y = arctan x = tan 1 x iff tan y = x < x < π 2 < y < π 2 y = arccot x = cot 1 x iff cot y = x < x < 0 < y < π y = arcsec x = sec 1 x iff sec y = x x 1 0 y π, y π 2 y = arccsc x = csc 1 x iff csc y = x x 1 π 2 y π 2, y 0

3 5.7. Inverse Trigonometric Functions: Differentiation Example 1. Evaluate each value. (a) arcsin 1 2 (b) arccos 0 (c) arctan 3 () arcsin(0.3) Example 2. Solve the equation. arctan(2x 3) = π 4 Example 3. (a) Given y = arcsin x, where 0 < y < π/2, fin cos y. (b) Given y = arcsec( 5/2), fin tan y.

4 5.7. Inverse Trigonometric Functions: Differentiation Theorem 5.18 Derivatives of Inverse Trigonometric Functions. Let u be a ifferentiable function of x. x arcsin u = u 1 u 2 x arctan u = u x 1 + u 2 x x arcsec u = u u u 2 1 arccos u = u arccot u = u 1 u u 2 x arccsc u = u u u 2 1

5 5.7. Inverse Trigonometric Functions: Differentiation Example 4. Fin each erivative (a) [arcsin 2x (b) arctan(3x) (c) arcsin x () x x x x arcsec(e2x ) Example 5. Differentiate the following function an simplify the answer. y = arcsin x + x 1 x 2 Example 6. Analyze the graph of the function. y = arctan x 2.

6 5.8. Inverse Trigonometric Functions: Integration Theorem 5.18 Let u be a ifferentiable function of x. x arcsin u = u 1 u 2 arctan u = u x 1 + u 2 x arcsec u = u u u 2 1 Theorem 5.19 Let u be a ifferentiable function of x, an let a > 0. 1 a 2 u 2 u = arcsin u a + C 1 a 2 + u 2 u = 1 a arctan u a + C u 2 1 u 2 a 2 u = 1 a arcsec u a + C

7 5.8. Inverse Trigonometric Functions: Integration Example 1. Integrate each inefinite integral. 1 a 4 x x b x 2 x c 1 x 4x 2 9 x

8 5.8. Inverse Trigonometric Functions: Integration Example 2. Fin 1 e 2x 1 x.

9 5.8. Inverse Trigonometric Functions: Integration Example 3. Fin x x 2 x.

10 5.8. Inverse Trigonometric Functions: Integration Example 4. Fin 1 x 2 4x + 7 x.

11 5.8. Inverse Trigonometric Functions: Integration Example 5. Fin 1 3x x x. 2 Then fin the area of the region boune by the graph of 1 f x = 3x x 2 The x-axis, an the lines x = 3 2 an x = 9 4.

12 Definitions of the Hyperbolic Functions sinh x = ex e x csch x = 1 2 sinh x, x 0 cosh x = ex + e x sech x = 1 2 cosh x sinh x tanh x = coth x = 1 cosh x tanh x, x 0 Remark. The notation sinh x is rea as the hyperbolic sine of x, cosh x as the hyperbolic cosine of x, an so on.

13 Note that Hyperbolic functions are not perioic

14 Hyperbolic Ientities cosh 2 x sinh 2 x = 1 tanh 2 x + sech 2 x = 1 cosh 2 x csch 2 x = 1 sinh cosh 2x x = 2 sinh 2x = 2 sinh x cosh x sinh(x + y) = sinh x cosh y + cosh x sinh y sinh(x y) = sinh x cosh y cosh x sinh y cosh(x + y) = cosh x cosh y + sinh x sinh y cosh(x y) = cosh x cosh y sinh x sinh y cosh cosh 2x x = 2 cosh 2x = cosh 2 x + sinh 2 x

15 Theorem 5.20 Differentiation an Integration of Hyperbolic Functions Let u be a ifferentiable function of x.. sinh u = cosh u u x cosh u = sinh u u x x tanh u = sech2 u u x coth u = csch2 u u sech u = sech u tanh u u x csch u = csch u coth u u x cosh u u = sinh u + C sinh u u = cosh u + C sech 2 u u = tanh u + C csch 2 u u = coth u + C sech u tanh u u = sech u + C csch u coth u u = csch u + C

16 Example 1. Fin a x sinh x2 3 b c x x ln cosh x x sinh x cosh x [ x 1 cosh x sinh x] x

17 Example 2. Fin the relative extrema of f x = x 1 cosh x sinh x. Sketch the graph as well.

18 Example 3. Power cables are suspene between two towers, forming the catenary shown in the Figure. The equation for this catenary is f x = a cosh x a. The istance between the two towers is 2b. Fin the slope of the catenary at the point where the cable meets the right-han tower. y b b b x

19 Example 4. Fin cosh 2x sinh 2 2x x.

20 Inverse Hyperbolic Functions Note that four (sinh x, tanh x, csch x, coth x) of the six hyperbolic functions are one-to-one, but cosh x an sech x are not one-to-one. By restricting the omain we can make cosh x an sech x one-to-one.

21 Theorem 5.21 Inverse Hyperbolic Functions Inverse Function Domain sinh 1 x = ln(x + x 2 + 1) (, ) cosh 1 x = ln(x + x 2 1) [1, ) tanh 1 x = 1 2 coth 1 x = x ln 1 x x + 1 ln x 1 sech 1 x = ln x2 x ( 1,1), 1 (1, ) (0, 1] csch 1 x = ln 1 x x2 x, 0 (0, )

22 Theorem 5.22 Differentiation an Integration Involving Inverse Hyperbolic Functions Let u be a ifferentiable function of x. x sinh 1 u = u x cosh 1 u = u 2 1 x tanh 1 u = u 1 u 2 x cosh 1 u = u 1 u 2 x sech 1 x = u u u 1 u 2 x csch 1 x = u 1 + u 2 1 u 2 ± a u = ln u + 2 u2 ± a 2 + C 1 a 2 u 2 u = 1 a + u ln 2a a u + C u 1 u a 2 ± u u = 1 2 a ln a + a2 ± u2 + C u u

23 Example 6. Fin x sinh 1 2x x [tanh 1 x 3 ].

24 Example 7. Fin 1 x 4 9x x x 2 x

25