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Derivatives Using Logarithms Worksheet

Derivatives Using Logarithms Worksheet
  • Page 1
 1.  
If y = (x + 17)8x, then find dydx.
a.
8xx + 17
b.
1(x+17)8[8xx + 17 + 8ln(x + 17)]
c.
(x + 17)8x[8xx + 17 + 8ln(x + 17)]
d.
8log (x + 17)
e.
(x + 17)8x[8xx + 17 - 8ln(x + 17)]


Solution:

y = (x + 17)8x

ln y = 8x ln (x + 17)
[Apply logarithm on both sides.]

1ydydx = 8xx + 17 + 8ln(x + 17)
[Differentiate both sides with respect to x.]

dydx = (x + 17)8x[8xx + 17 + 8ln(x + 17)]
[Substitute y = (x + 17)8x.]


Correct answer : (3)
 2.  
If y = (sin x)- x, then find dydx.

a.
II
b.
III
c.
V
d.
I
e.
IV


Solution:

y = (sin x)- x
[Write the function.]

ln y = - x ln (sin x)
[Take logarithm on both the sides.]

Differentiate y with respect to x.

(1y)dydx = - x (1sin x) (cos x) + ln (sin x) (- 1)
[Find dydx.]

dydx = - y [x cot x + ln (sin x)]
[Multiply by y on both the sides.]

= - (sin x)- x [x cot x + ln (sin x)]
[Replace y with (sin x)- x.]

Therefore dydx = - (sin x)- x [x cot x + ln (sin x)]


Correct answer : (4)
 3.  
If y = (3x2+5)1x, then find dydx.

a.
III
b.
IV
c.
II
d.
I
e.
V


Solution:

y = (3x2+5)1x
[Write the function.]

ln (y) = (1x) ln (3x2+5)
[Take logarithm on both the sides.]

(1y)dydx = (1x) (13x2+5) (6x)+ln (3x2+5) (-1x2)
[Find dydx.]

dydx = y[63x2+5-1x2 ln (3x2+5)]
[Multiply by y on both the sides.]

= (3x2+5)1x[63x2+5-1x2 ln (3x2+5)]
[Replace y with (3x2+5)1x.]

Therefore, dydx = (3x2+5)1x[63x2+5-1x2 ln (3x2+5)]


Correct answer : (1)
 4.  
Find dydx, if y = (2-x)x.


a.
IV
b.
I
c.
V
d.
III
e.
II


Solution:

y = (2-x)x
[Write the function.]

ln y = x ln (2-x)
[Take logarithm on both the sides.]

(1y)dydx = x (12-x) (- 1)+ln (2-x) (12x)
[Find dydx.]

dydx = y [ln (2-x)2x-x2-x]
[Multiply by y on both the sides.]

dydx = (2-x)x [ln (2-x)2x-x2-x]
[Replace y with (2-x)x.]


Correct answer : (4)
 5.  
If y = (1+5x)5x, then find dydx.

a.
II
b.
I
c.
IV
d.
V
e.
III


Solution:

y = (1+5x)5x
[Write the function.]

ln (y) = 5x ln (1+5x)
[Take logarithm on both the sides.]

(1y)dydx = 5x (11+5x)(5)+ln (1+5x)(5)
[Find dydx.]

dydx = 5y[5x1+5x+ln (1+5x)]
[Multiply by y on both the sides.]

= 5(1+5x)5x [5x1+5x+ln (1+5x)]
[Replace y with (1+5x)5x.]

Therefore, dydx = 5(1+5x)5x [5x1+5x+ln (1+5x)]


Correct answer : (5)
 6.  
If a function is defined by y = xcos 3x, then find y ′.


a.
II
b.
I
c.
V
d.
IV
e.
III


Solution:

y = xcos 3x
[Write the function.]

ln y = cos 3x (ln x)
[Take logarithm on both the sides.]

(1y) y ′ = cos 3x (1x) + (ln x) ( - 3sin 3x)
[Find y ′.]

y ′ = y[cos 3xx - 3sin 3x (ln x)]
[Multiply by y on both the sides.]

= xcos 3x[cos 3xx - 3sin 3x (ln x)]
[Replace y with xcos 3x.]

Therefore, y ′ = xcos 3x[- 3sin 3x (ln x) + cos 3xx]


Correct answer : (1)
 7.  
Find dydx, if y = x3 sec 5x.

a.
V
b.
I
c.
III
d.
II
e.
IV


Solution:

y = x3 sec 5x
[Write the function.]

ln y = 1 / 2 ln (x3 sec 5x)
[Take logarithm on both the sides.]

= 1 / 2 [ln (x3) + ln (sec 5x)]
[Use ln MN = ln M + ln N]

= 1 / 2 [3 ln (x) + ln (sec 5x)]
[Use log am = mlog a]

(1y)dydx = 1 / 2 [3 (1x)+1sec 5x (5sec 5x tan 5x)]
[Find dydx.]

dydx = y[32x+52 tan 5x]
[Multiply by y on both the sides.]

= x3 sec 5x[32x+52 tan 5x]
[Replace y with x3 sec 5x.]

Therefore, dydx = x3 sec 5x[32x+52 tan 5x]


Correct answer : (2)
 8.  
Find the derivative of y = xx e3x + 4.


a.
I
b.
III
c.
IV
d.
V
e.
II


Solution:

y = xx e3x + 4
[Write the function.]

ln y = x (ln x)+(3x+4) (ln e)
[Take logarithm on both the sides.]

(1y) dydx = [x (1x) + (ln x) 12x] + 3
[Find dydx.]

dydx = y[1x+ln x2x + 3]
[Multiply by y on both the sides.]

= xx e3x + 4 [1x+ln x2x + 3]
[Replace y with xx e3x + 4.]

Therefore, the derivative of y = xx e3x + 4 is dydx = xx e3x + 4 [1x+ln x2x + 3]


Correct answer : (1)
 9.  
Find dydx, if y = (x2-9)10(2x+4)6sin x.


a.
I
b.
III
c.
V
d.
IV
e.
II


Solution:

y = (x2-9)10(2x+4)6sin x
[Write the function.]

ln y = 10 [ln (x2 - 9)] + 6 [ln (2x + 4)] - [ln (sin x)]
[Take logarithm on both the sides.]

(1y) dydx = 10x2-9 (2x)+62x+4(2)-1sin x(cos x)
[Find dydx.]

dydx = y(20xx2-9+6x+2-cos xsin x)
[Mulitply by y on both the sides.]

= (x2-9)10(2x+4)6sin x (20xx2-9+6x+2-cos xsin x)
[Replcae y with (x2-9)10(2x+4)6sin x.]

Therefore, dydx = (x2-9)10(2x+4)6sin x (20xx2-9+6x+2-cot x)


Correct answer : (1)
 10.  
If y = (3x + 2)ln x, then find y ′.


a.
V
b.
III
c.
IV
d.
I
e.
II


Solution:

y = (3x + 2)ln x
[Write the function.]

ln y = (ln x) [ln (3x + 2)]
[Take logarithm on both the sides.]

1y y ′ = (ln x) (13x+2) (3) + [ln (3x +2)] (1x)
[Find dydx.]

y ′ = y[ln (3x+2)x+3ln x3x+2]
[Mulitply by y on both the sides.]

y ′ = (3x + 2)ln x [ln (3x+2)x+3ln x3x+2]
[Replace y with (3x + 2)ln x.]


Correct answer : (1)

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