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How do you differentiate #e^((x^2-x)^2) # using the chain rule?

1 Answer

Shwetank Mauria

Jun 19, 2016

#(df)/(dx)=2x(x-1)(2x-1)e^((x^2-x)^2)#

Explanation:

Here #f(x)=e^(g(x))#, where #g(x)=(h(x))^2# and #h(x)=x^2-x#

According to chain rule #(df)/(dx)=(df)/(dg)xx(dg)/(dh)xx(dh)/(dx)#

Hence #(df)/(dx)=e^((x^2-x)^2)xx2(x^2-x)xx(2x-1)#

= #e^((x^2-x)^2)xx2x(x-1)xx(2x-1)#

= #2x(x-1)(2x-1)e^((x^2-x)^2)#

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