# Math Help - Studyin for final tommorow (chain rule/derivative)

1. ## Studyin for final tommorow (chain rule/derivative)

i m pretty good on finding most derivatives.. but i m stuck on this one:

y = cos(cos(cosx))

can some1 show me how to find the derivative to this? i know the trick is using the chain rule. please simplyify the steps cause i sumtimes dont understand.

2. $y=\cos (\cos (\cos x))\implies y'=-\sin (\cos (\cos x))\cdot \left( \cos (\cos x) \right)'.$

Proceed from there, it's just a matter of chain rule.

3. Originally Posted by Krizalid
$y=\cos (\cos (\cos x))\implies y'=-\sin (\cos (\cos x))\cdot \left( \cos (\cos x) \right)'.$

Proceed from there, it's just a matter of chain rule.
hm not quite understanding

4. Originally Posted by Legendsn3verdie
hm not quite understanding
The extended chain rule states the following
$\frac{d}{dx}\bigg[f(g(h(x)))\bigg]=f'(g(h(x)))\cdot{g'(h(x))}\cdot{h'(x)}$
So seeing that $f(x),g(x),h(x)$ all equal cos(x)

We can see that $\frac{d}{dx}\bigg[\cos(\cos(\cos(x)))\bigg]=-\sin(\cos(\cos(x)))\cdot{-\sin(\cos(x))}\cdot{-\sin(x)}$

Does this make sense?

If you are trying to practice the chain rule try these on for size

$\text{Find}\text{ }\frac{d}{dx}\bigg[\ln(\cos(x^2))\bigg]$

$\text{Find}\text{ }\frac{d}{dx}\bigg[\cos(\sqrt[3]{1-\tan(x)})\bigg]$

and finally

$\text{Find}\text{ }\frac{d}{dx}\bigg[\ln\bigg(\frac{5x}{\cos(arcsin(x))}\bigg)\bigg]$

5. Hello,

I'm not sure showing him the direct formula would help lol

Originally Posted by Legendsn3verdie
hm not quite understanding
Ok, you should know the chain rule : $(u(v(x))'=v'(x) \cdot u'(v(x))$

Here, $u(t)=\cos t$ ( $u'(t)=-\sin t$) and $v(x)=\cos(\cos(x))=t$

$\cos(\cos(\cos(x)))=u(v(x))$

-> The derivative of $\cos(\cos(\cos(x)))$ is $v'(x) \cdot u'(v(x))=\bigg(\cos(\cos(x))\bigg)' \cdot \bigg(- \sin (\cos(\cos(x)))\bigg)$

Does it make sense to you ? :x