﻿ Exponential Functions Worksheet | Problems & Solutions

# Exponential Functions Worksheet

Exponential Functions Worksheet
• Page 1
1.
Find the derivative of $f$($x$) = sin 3$x$cos 4$x$ + 3$e$2$x$ + 6$x$3
 a. 18 b. 18 c. 18 d. 18

#### Solution:

f(x) = 3e2x+6x3+sin 3xcos 4x

f′(x) = ddx[3e2x+6x3+sin 3xcos 4x]
[Differentiate with respect to x on both sides.]

= 18x2+6e2x+3cos 3xcos 4x-4sin 3xsin 4x

2.
Find the derivative of $h$($x$) = .
 a. (- 4cot 2$x$ b. (4cot 2$x$ 6) c. (- cot 2$x$ d. (

#### Solution:

h(x) = ecot22x + 2x3

h′(x) = ddx[ecot2 2x + 2x3]
[Differentiate with respect to x on both sides.]

= ecot22x + 2x3ddx(cot22x + 2x3)
[Use dydx(ekx) = kekx.]

= (- 4cot 2xcosec22x + 6x2)ecot22x + 2x3

3.
Find the derivative of $g$($x$) = .
 a. b. - 3 c. d.

#### Solution:

g(x) = e37+cos 6x

g′(x) = ddx[e37+cos 6x]
[Differentiate with respect to x on both sides.]

= e37+cos 6xddx(37+cos 6x)

= - 3  sin 6x37+cos 6xe37+cos 6x
[Use dydx(kx) =k2kx.]

4.
Find the derivative of $g$($x$) = ($e$4$x$ + $e$5$x$ + $e$6$x$)19$x$.
 a. 19(${e}^{4x}\left(1+4x\right)+{e}^{6x}\left(1+6x\right)$) b. (${e}^{4x}\left(1+4x\right)-{e}^{5x}\left(1+5x\right)$) c. 19(${e}^{4x}\left(1+4x\right)+{e}^{5x}\left(1+5x\right)+{e}^{6x}\left(1+6x\right)$) d. (${e}^{4x}\left(1-4x\right)+{e}^{5x}\left(1-5x\right)\left(1+6x\right)$)

#### Solution:

g(x) = (e4x+e5x+e6x)19x

g′(x) = (e4x+e5x+e6x)ddx(19x)+19xddx(e4x+e5x+e6x)
[Use the product rule.]

= 19(e4x+e5x+e6x)+19x(4e4x+5e5x+6e6x)

= 19(e4x[1+4x]+e5x[1+5x]+e6x[1+6x])
[Take out 19 as common factor.]

5.
Find the derivative of $h$($x$) = + .
 a. $\frac{7}{2}$ b. $\frac{7}{2}$ c. - 3 d. $\frac{7}{2}$. + 3

#### Solution:

h(x) = ecos 7x + esin 6x

h′(x) = ddx[ ecos 7x] + ddx[esin 6x]

= ecos 7x.ddx[cos 7x] + esin 6x.ddx[sin 6x]

= 7 / 2- sin 7xcos 7x.ecos 7x + 3 cos 6xsin 6x.esin 6x

6.
Find the derivative of $f$($x$) = 29${e}^{{e}^{2x}}$.
 a. 58${e}^{{e}^{2x}}$ b. 58${e}^{2x}{e}^{{e}^{2x}}$ c. 29${e}^{2x}{e}^{{e}^{2x}}$ d. 2${e}^{2x}{e}^{{e}^{2x}}$

#### Solution:

f(x) = 29ee2x

f′(x) = ddx(29ee2x)
[Differentiate with respect to x on both sides.]

= 29e2x(2)ee2x
[dydx(ekx) = kekx.]

= 58e2xee2x
[58 = 29 × 2.]

7.
Find the derivative of $y$ = 40$e$- 5$x$ cos 6$x$.
 a. - $e$- 5$x$ [240sin 6$x$ + cos 6$x$] b. - $e$- 5$x$ [240sin 6$x$ - 200cos 6$x$] c. - $e$- 5$x$ [240sin 6$x$ + 200cos 6$x$] d. - $e$ 5$x$ [240sin 6$x$ + 200cos 6$x$]

#### Solution:

y = 40e - 5x cos 6x

dydx = 40ddx[ e - 5x cos 6x]
[Differentiate with respect to x on both sides.]

= 40e- 5x ddx(cos 6x) + 40(cos 6x) ddx( e- 5x)
[Use product rule.]

= - 240e- 5xsin 6x - 200e- 5x cos 6x
[Use cos kx = - ksin kx , dydx(ekx) = kekx.]

= - e- 5x [240sin 6x + 200cos 6x]

8.
Find $\frac{dy}{dx}$ if $y$ = 40$e$5$x$ log sin 4$x$.
 a. 40$e$$x$ [4tan 4$x$ + 5log(sin 4$x$)] b. 40$e$5$x$ [4cot 4$x$ + 5log(sin 4$x$)] c. 40$e$$x$ [4cot 4$x$ - 5log(cos 4$x$)] d. 40$e$5$x$ [4tan 4$x$ - 5log(cos 4$x$)]

#### Solution:

y = 40e5xlog sin 4x.

dydx = ddx[40e5x log(sin 4x)]
[Differentiate both sides with respect to x.]

= 40[e5x ddx(log(sin 4x)) + log(sin 4x)ddx(e5x)]
[Use product rule.]

= 40[4e5x cot 4x + 5e5x log(sin 4x)]
[Use log(sin x) = cot x.]

= 40e5x [4cot 4x + 5log(sin 4x)]

9.
Find the derivative of $f$($x$) = 46sin2(${e}^{4x}$).
 a. 184sin 2 b. 184${e}^{4x}$sin 2${e}^{4x}$ c. 184${e}^{4x}$sin d. 184${e}^{4x}$cos ${e}^{4x}$

#### Solution:

f(x) = 46sin2(e4x)

f′(x) = ddx[ 46sin2(e4x)]
[Differentiate with respect to x on both sides.]

= 92sin(e4x)cos(e4x)ddx(e4x)
[Use chain rule.]

= 368e4xsin(e4x)cos(e4x)
[Use chain rule.]

= 184e4x [2sin(e4x)cos(e4x)]
[Use 368 = 2×184.]

= 184e4xsin 2e4x
[Use 2sin θcos θ = sin 2θ.]

10.
Find the derivative of $h$($x$) = 8$x$2${e}^{3x}$.
 a. 8$x$${e}^{3x}$(3$x$ + 2) b. $x$${e}^{3x}$($x$ + 2) c. $x$${e}^{3x}$($x$² - 2) d. ${e}^{3x}$($x$² + 16)

#### Solution:

h(x) = 8x2e3x

h′(x) = ddx[8x2e3x]
[Differentiate with respect to x on both sides.]

= ddx(8x2e3x)
[Use product rule.]

= [8x2ddx(e3x) + e3xddx(8x2)]
[Use chain rule.]

= [8x2(3e3x) + e3x16x]
[Use dydx(ekx) = kekx.]

= 8xe3x[3x + 2]
[Take out 8xe3x as common factor.]