# How do you differentiate  f(x)=(ln(-x+3))^2 using the chain rule.?

Nov 4, 2016

The answer is $= \frac{- 2 \ln \left(- x + 3\right)}{- x + 3}$

#### Explanation:

We apply the chain rule.
$\left({x}^{n}\right) ' = n {x}^{n - 1}$
$\left(\ln x\right) ' = \frac{1}{x}$
$\left(a x + b\right) ' = a$

$\left({\left(\ln \left(- x + 3\right)\right)}^{2}\right) ' = 2 \ln \left(- x + 3\right) \cdot \left(\ln \left(- x + 3\right)\right) ' \cdot \left(- x + 3\right) '$

$= 2 \ln \left(- x + 3\right) \cdot \left(\frac{1}{- x + 3}\right) \left(- 1\right)$

$= \frac{- 2 \ln \left(- x + 3\right)}{- x + 3}$