How do you differentiate $f(x)=(1/(x-3)^2)^2$ using the chain rule?

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Answer 1 · 2016-02-24T00:58:16

Rewrite it as $f(x) = (x-3)^(-4)$

Apply the chain rule: $d/dx(u^-4) = -4u^-5 (du)/dx$

So $f'(x) = -4(x-3)^-5 d/dx(x-3)$

$= -4(x-3)^-5$

You may prefer to write the answer as $f'(x) = (-4)/(x-3)^5$

How do you differentiate f(x)=(1/(x-3)^2)^2f(x)=(1/(x-3)^2)^2 using the chain rule?