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

Feb 24, 2016

Rewrite it as $f \left(x\right) = {\left(x - 3\right)}^{- 4}$

Explanation:

Apply the chain rule: $\frac{d}{\mathrm{dx}} \left({u}^{-} 4\right) = - 4 {u}^{-} 5 \frac{\mathrm{du}}{\mathrm{dx}}$

So $f ' \left(x\right) = - 4 {\left(x - 3\right)}^{-} 5 \frac{d}{\mathrm{dx}} \left(x - 3\right)$

$= - 4 {\left(x - 3\right)}^{-} 5$

You may prefer to write the answer as $f ' \left(x\right) = \frac{- 4}{x - 3} ^ 5$