CHAIN RULE EXAMPLES WITH SOLUTIONS

About "Chain Rule Examples With Solutions"

Chain Rule Examples With Solutions :

Here we are going to see how we use chain rule in differentiation.

Chain Rule - Examples

Question 1 :

Differentiate f(x)  =  x/√(7 - 3x)

Solution :

u  =  x

u'  =  1

v  =  √(7 - 3x)

v'  =  1/2√(7 - 3x)(-3)  ==>  -3/2√(7 - 3x)==>-3/2√(7 - 3x)

f'(x)  =  [√(7 - 3x)(1) - x(-3/2√(7 - 3x))]/(√(7 - 3x))2

f'(x)  =  [√(7 - 3x) + (3x/2√(7 - 3x))]/(√(7 - 3x))2

f'(x)  =  [2(7 - 3x) + 3x)/2√(7 - 3x))]/(7 - 3x)

f'(x)  =  (14-6x+3x)/(2(7-3x)√(7-3x))

f'(x)  =  (14-3x)/[2(7-3x)√(7-3x)]

Question 2 :

Differentiate y = tan(cos x)

Solution :

y = tan(cos x)

Let u = cos x

Differentiate the function "u" with respect to "x"

du/dx  =  -sin x

y = tan u

Differentiate the function "y" with respect to "x".

dy/dx  = sec2 u (du/dx)

dy/dx  = sec2 (cos x)(-sin x)

dy/dx  =  -sin x sec2 (cos x)

Question 3 :

Differentiate y = sin2x/cos x

Solution :

u = sin2x  ==>  u'  =  2 sin x cos x

v  =  cos x  ==>  v'  =  - sinx

dy/dx  =  (cos x(2 sin x cos x) - sin2x (- sinx)) / (cos2x)

dy/dx  =  (2 sin x cos2 x + sin3x) / (cos2x)

dy/dx  =  2 sin x + (sin3x / cos2x)

=  2 sin x + tan2x sin x

=  sin x (2 + tan2x)

dy/dx  =  sin x (1 + sec2x)

Question 4 :

Differentiate y = 5-1/x

Solution :

Let u  =  -1/x

du/dx  =  -1/x2

y = 5u

dy/dx  =  5u (log 5) (du/dx)

=  5-1/x (log 5) ( -1/x2)

dy/dx   =  (-5-1/x log 5)/x2

Question 4 :

Differentiate y = √(1 + 2 tan x)

Solution :

Let u  =  (1 + 2 tan x)

du/dx  =  0 + 2 sec2x  ==> 2 sec2x

y =  √u

dy/dx  =  (1/2√u) (du/dx)

dy/dx  =  (1/2√(1 + 2 tan x) )(2 sec2x)

dy/dx  =  (sec2x/√(1 + 2 tan x))

Question 5 :

Differentiate y = sin3x + cos3x

Solution :

dy/dx  =  3 sin2x(cos x) + 3 cos2x(-sin x)

dy/dx  =  3 sinx cos x (sin x - cos x)

Question 6 :

Differentiate y = sin2 (cos kx)

Solution :

Let u = cos kx

Differentiate the function "u" with respect to "x"

du/dx  =  -sin kx (k)  ==>  - k sin kx

y = sinu

Differentiate the function "y" with respect to "x"

dy/dx  =  2 sin u cos u (du/dx)

=  sin 2u (du/dx)

=  sin (2 cos kx) (-k sin kx)

dy/dx  =  -k sin kx sin (2 cos kx)

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