# HOW TO USE THE CHAIN RULE TO DIFFERENTIATE

## About "How to Use the Chain Rule to Differentiate"

How to Use the Chain Rule to Differentiate :

Here we are going to see how we use chain rule to differentiate.

## Chain Rule to Differentiate - Examples

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.

∜(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)

After having gone through the stuff given above, we hope that the students would have understood, "How to Use the Chain Rule to Differentiate"

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