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

Apr 9, 2016

$f ' \left(x\right) = 4 {x}^{3} + 12 x$

#### Explanation:

The equation is in a ridiculous form above. You can cancel out the $e$ and the power $\ln$, leaving just

${e}^{\ln {\left({x}^{2} + 3\right)}^{2}} = {\left({x}^{2} + 3\right)}^{2}$

The chain rule says that if

$f \left(x\right) = g \left(h \left(x\right)\right)$

then

$f ' \left(x\right) = h ' \left(x\right) \cdot g ' \left(h\right)$.

Substituting $h \left(x\right) = {x}^{2} + 3$ and $g \left(h\right) = {h}^{2}$, then

$h ' \left(x\right) = 2 x$

$g ' \left(h\right) = 2 h = 2 \left({x}^{2} + 3\right) = 2 {x}^{2} + 6$

Multiplying these together according to the chain rule,

$h ' \left(x\right) \cdot g ' \left(h\right) = 2 x \cdot \left(2 {x}^{2} + 6\right) = 4 {x}^{3} + 12 x$

If you want to work it out another way, expand the brackets to get ${x}^{4} + 6 {x}^{2} + 9$ and differentiate with the usual method.