# How do you differentiate f(x)=ln(3(e^(sin^2x))^4) using the chain rule?

Nov 9, 2015

I would start with simplifying this expression as it is before differentiating.

Remember two rules:
$\ln \left(a \cdot b\right) = \ln \left(a\right) + \ln \left(b\right)$
$\ln \left({a}^{b}\right) = b \cdot \ln \left(a\right)$

Using this rules, the expression can be simplifyed in quite a drastical way:

$\ln \left(3 {\left({e}^{{\sin}^{2} x}\right)}^{4}\right)$
$= \ln \left(3\right) + \ln \left({\left({e}^{{\sin}^{2} x}\right)}^{4}\right)$
$= \ln \left(3\right) + 4 \cdot \ln \left({e}^{{\sin}^{2} x}\right)$
...remember that $\ln$ and $e$ are inverse functions...
$= \ln \left(3\right) + 4 \cdot {\sin}^{2} \left(x\right)$

Now, this is something that can be differentiated in a reasonably easy way - better than the former one anyway.

$f \left(x\right) = \ln \left(3\right) + 4 \cdot {\sin}^{2} \left(x\right) = \ln \left(3\right) + 4 \cdot {u}^{2} \left(x\right)$
where $u \left(x\right) = \sin \left(x\right)$.

As you know, $u ' \left(x\right) = \cos \left(x\right)$, so using the chain rule we get:

$f ' \left(x\right) = 0 + 4 \cdot 2 u \left(x\right) \cdot u ' \left(x\right)$
$= 4 \cdot 2 \cdot \sin \left(x\right) \cdot \cos \left(x\right)$
$= 8 \sin \left(x\right) \cos \left(x\right)$