# Differentiating functions with natural logs

• Nov 7th 2009, 03:45 PM
iheartphysics01
Differentiating functions with natural logs
Can anyone explain to me how to do this problem?

y=ln(cos(4x))
• Nov 7th 2009, 03:53 PM
Scott H
Welcome to the Math Help Forum! :)

The function $\displaystyle y=\ln\cos (4x)$ may be differentiated by applying the Chain Rule twice, knowing that

\displaystyle \begin{aligned} \frac{d}{dx}\ln x&=\frac{1}{x}\\ \frac{d}{dx}\cos x&=-\sin x\\ \frac{d}{dx}(kx)&=k. \end{aligned}

For reference, the Chain Rule states that

$\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))g'(x).$
• Nov 7th 2009, 03:53 PM
VonNemo19
Quote:

Originally Posted by iheartphysics01
Can anyone explain to me how to do this problem?

y=ln(cos(4x))

Sure!

$\displaystyle \frac{d}{dx}[\ln{u}]=\frac{u'}{u}$

In your problem, let $\displaystyle u=\cos(4x)$. This implies $\displaystyle u'=-4\sin{(4x)}$

So...

$\displaystyle \frac{d}{dx}[\ln{u}]=\frac{u'}{u}=\frac{-4\sin(4x)}{\cos(4x)}=-4\tan(4x)$
• Nov 7th 2009, 04:06 PM
iheartphysics01
ok thanks so much!(Happy)
i will probably be using this forum a lot!