HOW TO USE THE CHAIN RULE TO DIFFERENTIATE

Question 1 :

Differentiate y = (2x - 5)⁴ (8x² - 5)^-3

Solution :

To find dy/dx, first we have to use the product rule, in the product rule itself we have to use the chain rule. 

u  =  (2x - 5)⁴

u'  =  4(2x - 5)³ (2)  ==> 8 (2x - 5)³

v  =  (8x² - 5)^-3

v'  =  -3(8x² - 5)^-4 (16x)  ==>  -48x(8x² - 5)^-4

d(uv)  =  uv' + vu'

  =  (2x - 5)⁴(-48x(8x² - 5)^-4) + (8x² - 5)^-3(8 (2x - 5)³)

  =  -48x [(2x - 5)⁴/(8x² - 5)⁴] + 8[(2x - 5)³/(8x² - 5)³]

  =  8((2x - 5)³/(8x² - 5)⁴)[-6x(2x - 5) + (8x² - 5)]

  =  8((2x - 5)³/(8x² - 5)⁴)[-12x² + 30x + 8x² - 5]

  =  8((2x - 5)³/(8x² - 5)⁴)[-4x² + 30x - 5]

Question 2 :

Differentiate y = (x² + 1)∛(x² + 2)

Solution :

To find dy/dx, first we have to use the product rule, in the product rule itself we have to use the chain rule. 

u  =  (x² + 1)

u'  =  2x + 0  ==>  2x

v  =  ∛(x² + 2)  =  (x² + 2)^1/3

v'  =  (1/3)(x² + 2)^-2/3(2x)

d(uv)  =  uv' + vu'

  =  (x² + 1)(2x/3)(x² + 2)^-2/3 + (x² + 2)^1/3(2x)

  =  [2x(x² + 1)/3(x² + 2)^2/3] + 2x(x² + 2)^1/3

  =  (2x³ + 2x)+6x((x² + 2))/(x² + 2)^2/3

  =  (2x³ + 2x)+6x((x² + 2))/3(x² + 2)^2/3

  =  (2x³ + 2x + 6x³ + 12x)/3(x² + 2)^2/3

  =  (8x³ + 14x)/(x² + 2)^2/3

Question 3 :

Differentiate y = xe^(-x²)

Solution :

To find dy/dx, first we have to use the product rule, in the product rule itself we have to use the chain rule. 

u  =  x

u'  =  1

v  =  e^(-x²)   

v'  = e^(-x²) (-2x) ==> -2xe^(x²)

d(uv)  =  uv' + vu'

  =  x (-2xe^(-x²))+ e^(-x²) (1)

=  e^(-x²)[ -2x² + 1]

=  e^(-x²)[1 - 2x²]

Question 4 :

Differentiate y = ∜(t³ + 1)/(t³ - 1)

Solution :

To find dy/dx, we have to use the quotient rule, in the quotient rule itself we have to use the chain rule. 

y = ∜(t³ + 1)/(t³ - 1)

y⁴  =  (t³ + 1)/(t³ - 1)

Differentiate with respect to "t"

u  =  (t³ + 1) 

u'  =  3t²

v  =  (t³ - 1) 

v'  =  3t²

4y³ (dy/dt)  =  [(t³ - 1) 3t² - (t³ + 1) (3t²)]/(t³ - 1)²

dy/dt  =  [(t³ - 1) 3t² - (t³ + 1) (3t²)]/4y³(t³ - 1)²

dy/dt  =  (3t⁵ - 3t² - 3t⁵ - 3t²)/4((t³+1)/(t³-1))^3/4(t³-1)²

dy/dt  =  (- 3t² )/(2((t³+1)^3/4(t³-1)²/(t³-1))^3/4)

dy/dt  =  (- 3t² )/(2((t³+1)^3/4(t³-1))^2-3/4)

dy/dt  =  (- 3t² )/(2((t³+1)^3/4(t³-1))^5/4)

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.