Derivatives of Exponential Functions Worksheet

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) = e^{1 + 1 + 1 + .....+ 1}(1 + 2 + 3 + .....+ 20)
[Find g′(1).]

= 20(20+1)2e21 = 210e²¹

g′(1) = 210e²¹

Correct answer : (1)

f

(

q

) =

e^{(8 \sin^{2} 3 q + 12 \cos^{2} 3 q)}

, then find

f

′ (

t

	a.		$e^{(8 \sin^{2} 3 q + 12 \cos^{2} 3 q)}$
	b.		- 16sin 3 $q$ + 24cos 3 $q$
	c.		16 $\cos^{2} 3 q + 24 \sin^{2} 3 q$
	d.		- 12sin 6 $q$ $e^{(8 \sin^{2} 3 q + 12 \cos^{2} 3 q)}$
	e.		60 $e^{(8 \sin^{2} 3 q + 12 \cos^{2} 3 q)}$

Solution:

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

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

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

Correct answer : (4)

f

(

r

) =

\sqrt{4 e^{8 r} + 8 e^{r} + 12}

, then find

f

′ (

r

	a.		32 $e^{8 r} + 8 e^{r}$
	b.		$\frac{1}{2 \sqrt{4 e^{8 r} + 8 e^{r} + 12}}$
	c.		Does not exist
	d.		$\frac{32 e^{8 r} + 8 e^{r}}{2 \sqrt{4 e^{8 r} + 8 e^{r} + 12}}$
	e.		$\frac{8 e^{r}}{2 \sqrt{4 e^{8 r} + 8 e^{r} + 12}}$

Solution:

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

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

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

= (4) (8)e8r+8er24e8r+8er+12

f ′ (r) = 32e8r+8er24e8r+8er+12

Correct answer : (4)

h

(

m

) = tan (

e^{15 m}) + \sec (e^{15 m})

, then find

h

′ (

m

	a.		sec ( $e^{15 m}) (\sec (e^{15 m}) - \tan (e^{15 m}))$
	b.		15 $e^{15 m} \sec (e^{15 m}) (\sec (e^{15 m}) + \tan (e^{15 m}))$
	c.		Does not exist
	d.		sec ( $e^{15 m}) + \tan (e^{15 m})$
	e.		sec ( $e^{15 m}) (\sec (e^{15 m}) + \tan (e^{15 m}))$

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))

Correct answer : (2)

f

(

s

) =

e^{s} + \frac{e^{s^{2}}}{2} + \frac{e^{s^{3}}}{3} + . . . . + \frac{e^{s^{90}}}{90}

, then find

f

′(1).

	a.		90 $e$
	b.		1 + $\frac{1}{2} + \frac{1}{3} + . . . . + \frac{1}{90}$
	c.		89 $e$
	d.		90
	e.		$\frac{1}{2} + \frac{1}{3} + . . . . + \frac{1}{90}$

Solution:

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

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

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

f ′(s) = es+ses2+s2es3+....+s89es90

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

Correct answer : (1)

Find

f

′ (

m

), if

f

(

m

) = sin

\sqrt{8 e^{3 m} + 41}

	a.		( $\sqrt{8 e^{3 m} + 41}$ )
	b.		( $e^{3 m} \cos \sqrt{8 e^{3 m} + 41}$ )
	c.		12 ( $\frac{e^{3 m} \cos \sqrt{8 e^{3 m} + 41}}{\sqrt{8 e^{3 m} + 41}}$ )
	d.		$\frac{1}{2}$ cos 24 $e$ ^{3 $m$}
	e.		(cos $\sqrt{8 e^{3 m} + 41}$ )

Solution:

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

f ′ (m) = ddm (sin8e3m+41)
[Find f ′ (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)

f ′ (m) = 12 (e3m cos8e3m+418e3m+41)

Correct answer : (3)

f

(

x

) =

e^{(- 6 x + \tan 6 x)}

, then find

f

′ (

x

	a.		6 (sec 6 $x$ tan 6 $x$ - 1) $e^{(- 6 x + \tan 6 x)}$
	b.		6 sec² 6 $x$ $e^{(- 6 x + \tan 6 x)} - 6 e^{6 x}$
	c.		$e^{(- 6 x + \tan 6 x)}$
	d.		(tan² 6 $x$ - 6) $e^{(- 6 x + \tan 6 x)}$
	e.		6 tan² 6 $x$ $e^{(- 6 x + \tan 6 x)}$

Solution:

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

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

f ′ (x) = 6 tan² 6x e(-6x+tan 6x)

Correct answer : (5)

f

(

k

) =

e

^{9^$k$}, then find

f

′ (

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:

f (k) = e^{9^k}
[Write the function.]

f ′ (k) = ddk (e^{9^k})
[Find f ′ (k)]

= e^{9^k} ddk (9^k)
[Use the Chain Rule.]

= e^{9^k} 9^k ln 9

f ′ (k)= 9^k e^{9^k} ln 9

Correct answer : (3)

f

(

q

) = 3^{1 + $q$ ² + $q$ ⁴ + ....+ $q$ ²⁴,} then find

f

′(1).

	a.		3¹³ ln 3
	b.		(312)3¹³ ln 3
	c.		(156)3¹³ ln 3
	d.		(156)3¹³
	e.		3¹⁵⁶

Solution:

f (q) = 3^{1 + q² + q⁴ + ....+ q²⁴}
[Write the function.]

f ′(q) = ddq(3^{1 + q² + q⁴ + ....+ q²⁴})
[Find f ′(q).]

= (3^{1 + q² + q⁴ + ....+ q²⁴})(ln 3)(2q + 4q³ + ....+ 24q²³)
[Use the Chain Rule.]

f ′(q) = 3^{1 + q² + q⁴ + ....+ q²⁴}(2q + 4q³ + ....+ 24q²³)ln 3

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

f ′(1) = (156)3¹³ ln 3

Correct answer : (3)

10.

h

(

k

) =

\frac{16 k}{9 + e^{11 k}}

, then find

h

′ (

k

	a.		[144 + 16 $e^{11 k} (1 - 11 k)] {(9 + e^{11 k})}^{2}$
	b.		144 + 16 $e^{11 k}$ (1-11 $k$ )
	c.		9 + $e^{11 k}$
	d.		[144 + 16 $e^{11 k} (1 - 11 k)] {(9 + e^{11 k})}^{- 2}$
	e.		144 + 16 $e^{11 k}$ (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

Correct answer : (4)