# How do you find the derivative of f(x)=ln(2x^2+1)^5?

Dec 13, 2016

Chain Rule

There are three "layers" in this function.

(f(g(h(x)))

Going from:
outside
$\downarrow$
inside

$f \left(x\right) = {x}^{5}$
$f ' \left(x\right) = 5 {\left(x\right)}^{4}$

$g \left(x\right) = \ln \left(x\right)$
$g ' \left(x\right) = \frac{1}{x}$

$h \left(x\right) = \left(2 {x}^{2} + 1\right)$
$h ' \left(x\right) = 4 x$

#### Explanation:

To find the derivative, we need to apply this rule.

Thus:

$f ' \left(x\right) = \frac{5 \ln {\left(2 {x}^{2} + 1\right)}^{4}}{2 {x}^{2} + 1} \cdot 4 x = \frac{20 x \ln \left(2 {x}^{2} + 1\right)}{2 {x}^{2} + 1}$