Find the derivative of (x^2+1) ^3(x-1) ^2.

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Here we have to find the derivative of [(x^2+1) ^3]*[(x-1) ^2]

Let’s use the product rule first.

=> (x^2 +1) ^3 * d/dx(x-1) ^2 + d/dx ((x^2+1) ^3)*(x-1) ^2

Now we use the product rule for the individual terms

d/dx(x-1) ^2 = 2 (x-1) *1 = 2(x-1)

d/dx (x^2+1) ^3 = 3(x^2+1) ^2 * 2x = 6x(x^2+1) ^2

So we get

(x^2 +1)^3 * d/dx(x-1) ^2 + d/dx ((x^2+1) ^3)*(x-1) ^2

=> (x^2 +1) ^3 * 2(x-1) + 6x(x^2+1) ^2*(x-1) ^2

Simplifying now

=> 2(x-1) (x^2+1) ^3 + 6x (x-1) ^2 (x^2+1) ^2

=> 2(x^2 +1) ^2 (x-1) [x^2+1 + 3x(x-1)]

=> 2(x^2 +1) ^2 (x-1) [x^2+1 + 3x^2 – 3x]

=> 2(x^2 +1) ^2 (x-1) [4x^2-3x+1]

**Therefore we finally get 2(x^2 +1) ^2 (x-1) [4x^2-3x+1]**

y ==(x^2+1)^3(x-1)^2.

To find the derivative of a function y = { h(x)} ^g(x)}, we use

y' ={ (h(x))^g(x)}' = y {g'(x)lnf(x)+g(x)f'(x)/f(x)}.

Here, g(x) = 3(x-1)^2. g'(x) = 3*2(x-1) = 6(x-1).

h(x) = x^2+1. h'(x) = (x^2+1) = 2x.

Therefore ,

{(x^2+1)^3(x-1)^2}' = y {6(x-1)ln(x^2+1) + 3(x-1)^2*2x/(x^2+1)}

{(x^2+1)^3(x-1)^2}' = {(x^2+1)^3(x-1)^2}{6(x-1)ln(x^2+1) +6x(x-1)^2 /(x^2+1)}

{(x^2+1)^3(x-1)^2}' = {(x^2+1)^3(x-1)^2}{6(x-1)[ln(x^2+1) +6x(x-1) /(x^2+1)]}

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