EXAMPLE PROBLEMS IN DIFFERENTIATION USING CHAIN RULE

Question 1 :

Differentiate y = (1 + cos2x)6

Solution :

u  =  1 + cos2x

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

du/dx  =  0 + 2 cos x (-sin x)  ==>  -2 sin x cos x

du/dx  =  - sin 2x

y = u

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

dy/dx  =  6u5 (du/dx)

  =  6(1 + cos2x)(-sin 2x)

  =  -6 sin 2x (1 + cos2x)5

Question 2 :

Differentiate y = e3x/(1 + ex)

Solution :

u  =  e3x  ==>  u'  =  3e3x

v  =  1 + ex  ==>  v'  =  ex

dy/dx  =  [(1 + ex) (3e3x) - e3x (ex)]/(1 + ex)2

dy/dx  =  [3e3x + 3e4x - e4x]/(1 + ex)2

dy/dx  =  [3e3x + 2e4x]/(1 + ex)2

Question 3 :

Differentiate y = √(x+√x)

Solution :

y = √(x+√x)

y2 = x + √x

2y (dy/dx)  =  1 + 1/2√x

dy/dx  =  (1/2√(x+√x)) (1 + (1/2√x))

dy/dx  =  (1/2√(x+√x)) (2√x + 1)/2√x)

dy/dx  =  (2√x + 1)/4√x(√(x+√x))

Question 4 :

Differentiate y = excos x

Solution :

 y = excos x

Let u  =  x cos x

du/dx  =  x (-sinx ) + cos x (1)

du/dx  =  - x sin x + cos x 

y = eu

dy/dx  =  eu (du/dx)

dy/dx  =  ex cos x (- x sin x + cos x)

Question 5 :

Differentiate the following

Solution :

y2  =  x + √(x +√x)

2y (dy/dx)  =  1 + 1/2√(x +√x))

Question 6 :

Differentiate y = sin (tan (√sin x))

Solution :

dy/dx  =  cos (tan (√sin x)) sec2 (√sin x) (1/2√sin x) cos x

dy/dx  =  [cos (tan (√sin x)) sec2 (√sin x) cos x]/2√sin x

Question 7 :

Differentiate y = sin-1[(1 - x2)/ (1 + x2)]

Solution :

y = sin-1[(1 - x2)/ (1 + x2)]

Let u  =  (1 - x2)/(1 + x2)

du/dx  =  [(1 + x2)(-2x) - (1 - x2)(2x)]/(1 + x2)2

=  (-2x - 2x3 - 2x + 2x3) /(1 + x2)2

du/dx  =  -4x /(1 + x2)2

y = sin-1u

dy/dx  = 1/√(1 - u2) (du/dx)

= 1/√(1 -  ((1 - x2)/(1 + x2))2) [-4x /(1 + x2)2]

=  (1 + x2)/√(1 + x4 + 2x2 - 1 + 2x2 - x4)[-4x /(1 + x2)2]

=  ((1 + x2)/2x)[-4x /(1 + x2)2]

dy/dx  =  -2/(1 + x2)

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