DIFFERENTIATION PRACTICE QUESTIONS WITH ANSWERS

Find the derivatives of the following functions with respect to corresponding independent variables :

Question 1 :

Differentiate f(x) = x - 3 sinx

Solution :

f(x) = x - 3 sinx

f'(x)  =  1 - 3 cos x

Question 2 :

Differentiate y = sin x + cos x

Solution :

f(x) = sin x + cos x

f'(x)  =  cos x - sin x

Question 3 :

Differentiate f(x) = x sin x

Solution :

f(x) = x sin x

We have to use the product rule to find the derivative.

u  =  x ==>  u'  =  1

v  =  sin x ==>  v'  =  cos x

Product rule :

d(uv)  =  uv' + vu'

f'(x)  =  x(cos x) + sin x (1)

f'(x)  =  x cos x + sin x

Question 4 :

Differentiate y = cos x - 2 tan x

Solution :

f(x)  =  cos x - 2 tan x

f'(x)  =  -sin x - 2 sec² x

Question 5 :

Differentiate g(t) = t³cos t

Solution :

We have to use the product rule to find the derivative.

u  =  t³ ==>  u'  =  3t²

v  =  cos t ==>  v'  =  -sin t

f('x)  =  t³(-sin t) + cos t (3t²)

f('x)  =  -t³sin t + 3t²cos t

=  t² (3 cos t - t sin t)

Question 6 :

Differentiate g(t) = 4 sec t + tan t

Solution :

g(t) = 4 sec t + tan t

g'(t)  =  4 sec t tan t + sec² t

Question 7 :

Differentiate y = e^x sin x

Solution :

y = e^x sin x

u = e^x  ===> u'  =  e^x 

v = sin x ===> v'  =  cos x 

y'  =  e^x (cos x) + sin x(e^x)

y' =  e^x (cos x + sin x)

Question 8 :

Differentiate y  =  tan x / x

Solution :

y  =  tan x / x

u = tan x  ===> u'  =  sec² x 

v = x ===> v'  =  1

Quotient rule :

d(u/v)  =  (vu' - uv') / v²

dy/dx  =  (x sec² x  - tan x (1)) / x²

=  (x sec² x  - tan x) / x² 

Question 9 :

Differentiate y  =  sin x / (1 + cos x)

Solution :

y  =  sin x / (1 + cos x)

u = sin x ===> u'  =  cos x 

v = (1 + cos x) ===> v'  =  - sin x

Quotient rule :

d(u/v)  =  (vu' - uv') / v²

dy/dx  =  ((1 + cos x) cos x - sin x (-sin x)) / (1 + cos x)²

dy/dx  =  (cos x + cos² x + sin² x) / (1 + cos x)²

dy/dx  =  (1 + cos x) / (1 + cos x)²

dy/dx  =  1/(1 + cos x)

Question 10 :

Differentiate y  =  x / (sin x + cos x)

Solution :

y  =  x / (sin x + cos x)

u = x ===> u'  =  1 

v = (sin x + cos x) ===> v'  =  cos x - sin x

dy/dx  =  [(sinx+cosx) (1)-x(cosx-sinx)]/(sin x+cosx)²

dy/dx  =  [sinx + cosx - x cosx + xsinx)]/(sin x+cosx)²

dy/dx  =  [(1 + x) sinx + (1 - x) cosx]/(sin x+cosx)²

Question 11 :

If f(x) = (x - 1)(x² + 2)³, the f'(x) = 

a)  6x (x² + 2)²      b) 6x(x - 1) (x² + 2)²

c)  (x² + 2)² (x² + 3x - 1)    d)  (x² + 2)² (7x² - 6x + 2)

e)  -3(x - 1) (x² + 2)²

Solution :

f(x) = (x - 1)(x² + 2)³

Since these two terms are multiplied, we have to use product rule to find the derivative.

u = x - 1 and v = (x² + 2)³

u' = 1 and v' = 3(x² + 2)² (2x)

v' = 6x(x² + 2)²

f'(x) = (x - 1) 6x(x² + 2)² + (x² + 2)³ (1)

= 6x(x - 1) (x² + 2)² + (x² + 2)³

Factoring (x² + 2)², we get

= (x² + 2)²[6x(x - 1)  + (x² + 2)]

= (x² + 2)²[6x² - 6x + x² + 2]

= (x² + 2)²(7x² - 6x + 2)

So, option d is correct.

Question 12 :

If f(x) = e^(2/x), then f'(x) = 

a)  2 e^(2/x) ln x     b)  e^(2/x)     c) e^(-2/x²)

d)  (-2/x²) e^(2/x)     e)  -2x² e^(2/x)

Solution :

f(x) = e^(2/x) = e^(2x^-1)

f'(x) = e^(2/x) (-2x^-2)

= e^(2/x) (-2/x²)

So, option d is correct.

Question 13 :

In the xy-plane, the line x + y = k, where k is constant is tangent to the graph of y = x² + 3x + 1, what is the value of k ?

a)  -3    b)  -2    c)  -1    d)  0     e)  1

Solution :

y = x² + 3x + 1

dy/dx = 2x + 3(1) + 0

Slope of the curve at any point = 2x + 3 -----(1)

x + y = k

y = -x + k

Slope of the line = -1 -----(2)

(1) = (2)

2x + 3 = -1

2x = -1 - 3

2x = -4

x = -2

Let us find the point where the curve and tangent meets.

 y = k - x

y = x² + 3x + 1

Applying the value of y, we get

k - x = x² + 3x + 1

x² + 3x + 1 + x - k = 0

x² + 4x + 1 - k = 0

applying the value of x, we get

(-2)² + 4(-2) + 1 - k = 0

4 - 8 + 1 - k = 0

-4 + 1 - k = 0

-3 - k = 0

-k = 3

k = -3

So, option a is correct.

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