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Derivatives of Exponential Functions Worksheet

Derivatives of Exponential Functions Worksheet
  • Page 1
 1.  
If g(n) = e1+n+n2+...+n20, then find g′(1)
a.
210e21
b.
210e20
c.
e20
d.
420e21
e.
e21


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


Correct answer : (1)
 2.  
If (q) = e(8sin2 3q+12cos2 3q), then find ′ (t).
a.
e(8sin2 3q+12cos2 3q)
b.
- 16sin 3q + 24cos 3q
c.
16cos2 3q+24sin2 3q
d.
- 12sin 6q e(8sin2 3q+12cos2 3q)
e.
60e(8sin2 3q+12cos2 3q)


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)


Correct answer : (4)
 3.  
If (r) = 4e8r+8er+12, then find ′ (r).
a.
32e8r+8er
b.
124e8r+8er+12
c.
Does not exist
d.
32e8r+8er24e8r+8er+12
e.
8er24e8r+8er+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


Correct answer : (4)
 4.  
If h(m) = tan (e15m)+sec (e15m), then find h ′ (m).
a.
sec (e15m)(sec(e15m)-tan(e15m))
b.
15e15msec(e15m)(sec(e15m)+tan(e15m))
c.
Does not exist
d.
sec (e15m)+tan (e15m)
e.
sec (e15m)(sec(e15m)+tan(e15m))


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)
 5.  
If (s) = es+es22+es33+....+es9090, then find ′(1).
a.
90e
b.
1 + 12+13+ .... +190
c.
89e
d.
90
e.
12+13+ .... +190


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).]


Correct answer : (1)
 6.  
Find ′ (m), if (m) = sin 8e3m+41.
a.
(8e3m+41)
b.
(e3m cos8e3m+41)
c.
12 (e3m cos8e3m+418e3m+41)
d.
1 2 cos 24e3m
e.
(cos 8e3m+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)


Correct answer : (3)
 7.  
If (x) = e(- 6x+tan 6x), then find ′ (x).
a.
6 (sec 6x tan 6x - 1) e(-6x+tan 6x)
b.
6 sec2 6x e(-6x+tan 6x)-6e6x
c.
e(-6x+tan 6x)
d.
(tan2 6x - 6) e(-6x+tan 6x)
e.
6 tan2 6x e(-6x+tan 6x)


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)


Correct answer : (5)
 8.  
If (k) = e9k, then find ′ (k).
a.
k 9k -1 e9k
b.
9k e9k
c.
9k e9k ln 9
d.
9k e9k- 1
e.
e9k


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


Correct answer : (3)
 9.  
If (q) = 31 + q2 + q4 + ....+ q24, 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


Correct answer : (3)
 10.  
If h(k) = 16k9+e11k, then find h ′ (k).
a.
[144 + 16e11k(1-11k)](9+e11k) 2
b.
144 + 16e11k(1-11k)
c.
9 + e11k
d.
[144 + 16e11k(1-11k)](9+e11k)- 2
e.
144 + 16e11k(1+11k)


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)

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