﻿ Derivatives of Exponential Functions Worksheet | Problems & Solutions

# Derivatives of Exponential Functions Worksheet

Derivatives of Exponential Functions Worksheet
• Page 1
1.
If $g$($n$) = ${e}^{1+n+{n}^{2}+...+{n}^{20}}$, then find $g$′(1)
 a. 210$e$21 b. 210$e$20 c. $e$20 d. 420$e$21 e. $e$21

#### Solution:

g(n) = e1+n+n2+...+n20
[Write the function.]

g′(n) = ddn(e1+n+n2+...+n20)
[Find g′(n).]

= e1+n+n2+...+n20(1 + 2n + 3n2 +....+ 20n19)
[Use the Chain Rule.]

g′(n) = e1+n+n2 +...+n20(1 + 2n + 3n2 +...+ 20n19)

g′(1) = e1 + 1 + 1 + .....+ 1(1 + 2 + 3 + .....+ 20)
[Find g′(1).]

= 20(20+1)2e21 = 210e21

g′(1) = 210e21

2.
If ($q$) = , then find ′ ($t$).
 a. b. - 16sin 3$q$ + 24cos 3$q$ c. 16 d. - 12sin 6$q$ e. 60

#### Solution:

(q) = e(8sin2 3q+12cos2 3q)
[Write the function.]

′ (q) = ddq( e(8sin2 3q+12cos2 3q))
[Find ′ (q)]

= e(8sin2 3q+12cos2 3q)ddq (8sin2 3q+12cos2 3q)
[Use the Chain Rule.]

= e(8sin2 3q+12cos2 3q) ((8) (2) (sin 3q) (cos 3q) (3)+12 (2) (cos 3q) (- sin 3q) (3))
[Use the Chain Rule again.]

= e(8sin2 3q+12cos2 3q) (24sin 6q-36sin 6q)
[Use 2sin θ cos θ = sin 2θ]

= e(8sin2 3q+12cos2 3q) (- 12sin 6q)

= (- 12sin 6q) e(8sin2 3q+12cos2 3q)

′ (q) = - 12sin 6q e(8sin2 3q+12cos2 3q)

3.
If ($r$) = $\sqrt{4{e}^{8r}+8{e}^{r}+12}$, then find ′ ($r$).
 a. 32${e}^{8r}+8{e}^{r}$ b. $\frac{1}{2\sqrt{4{e}^{8r}+8{e}^{r}+12}}$ c. Does not exist d. $\frac{32{e}^{8r}+8{e}^{r}}{2\sqrt{4{e}^{8r}+8{e}^{r}+12}}$ e. $\frac{8{e}^{r}}{2\sqrt{4{e}^{8r}+8{e}^{r}+12}}$

#### Solution:

(r) = 4e8r+8er+12
[Write the function.]

′ (r) = ddr (4e8r+8er+12)
[Find ′ (r)]

= 124e8r+8er+12ddr (4e8r+8er+12)
[Use the Chain Rule.]

= (4) (8)e8r+8er24e8r+8er+12

′ (r) = 32e8r+8er24e8r+8er+12

4.
If $h$($m$) = tan (, then find $h$ ′ ($m$).
 a. sec (${e}^{15m}\right)\left(\mathrm{sec}\left({e}^{15m}\right)-\mathrm{tan}\left({e}^{15m}\right)\right)$ b. 15${e}^{15m}\mathrm{sec}\left({e}^{15m}\right)\left(\mathrm{sec}\left({e}^{15m}\right)+\mathrm{tan}\left({e}^{15m}\right)\right)$ c. Does not exist d. sec ( e. sec (${e}^{15m}\right)\left(\mathrm{sec}\left({e}^{15m}\right)+\mathrm{tan}\left({e}^{15m}\right)\right)$

#### Solution:

h(m) = tan (e15m)+sec (e15m)
[Write the function.]

h ′ (m) = ddm (tan (e15m)+sec (e15m))
[Find h′ (m)]

= sec2 (e15m) (e15m) (15)+sec (e15m) tan (e15m)(e15m) (15)
[Use the Sum Rule, the Chain Rule.]

= 15e15m sec (e15m)  (sec (e15m)+tan (e15m))
[Factor out 15e15m]

h ′ (m) = 15e15m sec (e15m)  (sec (e15m)+tan (e15m))

5.
If ($s$) = ${e}^{s}+\frac{{e}^{{s}^{2}}}{2}+\frac{{e}^{{s}^{3}}}{3}+....+\frac{{e}^{{s}^{90}}}{90}$, then find ′(1).
 a. 90$e$ b. 1 + c. 89$e$ d. 90 e.

#### Solution:

(s) = es+es22+es33+....+es9090
[Write the function.]

′(s) = dds(es+es22+es33+....+es9090)
[Find ′(s).]

= es+ses2+s2es3+....+s89es90
[Use the Sum Rule and the Chain Rule.]

′(s) = es+ses2+s2es3+....+s89es90

′(1) = e + e + e + ...... + e = 90e
[Find ′(1).]

6.
Find ′ ($m$), if ($m$) = sin $\sqrt{8{e}^{3m}+41}$.
 a. ($\sqrt{8{e}^{3m}+41}$) b. () c. 12 () d. $\frac{1}{2}$ cos 24$e$3$m$ e. (cos $\sqrt{8{e}^{3m}+41}$)

#### Solution:

(m) = sin 8e3m+41
[Write the function.]

′ (m) = ddm (sin8e3m+41)
[Find ′ (m)]

= cos 8e3m+41 ddm (8e3m+41)
[Use the Chain Rule.]

= (cos 8e3m+41) (128e3m+41) (8e3m) (3)
[Use the Chain Rule again.]

= 12 (e3m8e3m+41) (cos8e3m+41)

′ (m) = 12 (e3m cos8e3m+418e3m+41)

7.
If ($x$) = , then find ′ ($x$).
 a. 6 (sec 6$x$ tan 6$x$ - 1) b. 6 sec2 6$x$ c. d. (tan2 6$x$ - 6) e. 6 tan2 6$x$

#### Solution:

(x) = e(-6x+tan 6x)
[Write the function.]

′ (x) = ddx (e(-6x+tan 6x))
[Find ′ (x)]

= e(-6x+tan 6x)ddx (-6x+tan 6x)
[Use the Chain Rule.]

= e(-6x+tan 6x) (-6+6sec2 6x)
[Use the Chain Rule again.]

= (6 tan2 6x) e(-6x+tan 6x)

′ (x) = 6 tan2 6x e(-6x+tan 6x)

8.
If ($k$) = $e$9$k$, then find ′ ($k$).
 a. $k$ 9$k$ -1 $e$9$k$ b. 9$k$ $e$9$k$ c. 9$k$ $e$9$k$ ln 9 d. 9$k$ $e$9$k$- 1 e. $e$9$k$

#### Solution:

(k) = e9k
[Write the function.]

′ (k) = ddk (e9k)
[Find ′ (k)]

= e9k ddk (9k)
[Use the Chain Rule.]

= e9k 9k ln 9

′ (k)= 9k e9k ln 9

9.
If ($q$) = 31 + $q$2 + $q$4 + ....+ $q$24, then find ′(1).
 a. 313 ln 3 b. (312)313 ln 3 c. (156)313 ln 3 d. (156)313 e. 3156

#### Solution:

(q) = 31 + q2 + q4 + ....+ q24
[Write the function.]

′(q) = ddq(31 + q2 + q4 + ....+ q24)
[Find ′(q).]

= (31 + q2 + q4 + ....+ q24)(ln 3)(2q + 4q3 + ....+ 24q23)
[Use the Chain Rule.]

′(q) = 31 + q2 + q4 + ....+ q24(2q + 4q3 + ....+ 24q23)ln 3

′(1) = 313(2 + 4 + ....+ 24)ln 3
[Find ′(1).]

= (313)(2)(1 + 2 + 3 + ....+ 12)ln 3
[Factor out 2.]

= (313)(2)12(12 + 1)2ln 3
[Sum of the first n positive integers = n(n+1)2.]

= (156)313ln 3

′(1) = (156)313 ln 3

10.
If $h$($k$) = $\frac{16k}{9+{e}^{11k}}$, then find $h$ ′ ($k$).
 a. [144 + 16 b. 144 + 16${e}^{11k}$(1-11$k$) c. 9 + ${e}^{11k}$ d. [144 + 16 e. 144 + 16${e}^{11k}$(1+11$k$)

#### Solution:

h(k) = 16k9+e11k
[Write the function.]

h ′ (k) = ddk (16k9+e11k)
[Find h ′ (k)]

= (9+e11k)ddk (16k)-16kddk (9+e11k)(9+e11k)2
[Use the Chain Rule.]

= (9+e11k) (16)-16k(e11k) (11)(9+e11k)2

= 144+16e11k(1-11k)(9+e11k)2

h ′ (k) = 144+16e11k(1-11k)(9+e11k)2