Chain Rule Inside of Chain Rule

**DontDrinkAndDerive** · October 22nd 2009, 08:21 PM

I recently just took a Derivative Mastery Test...and although i passed, there was one problem that just stumped me and i could use some help so i dont make this mistake again.

Here's the problem:

Find Derivative of Arcsin(tan(3x^2))
and thats it...
Please and Thank you if you can give me a run through

**scorpion007** · October 22nd 2009, 08:32 PM

$f(x)=\arcsin(\tan(3x^2))$

$f'(x) = \frac{1}{\sqrt{1-\tan^2(3x^2)}}\cdot \sec^2(3x^2)\cdot 6x$ .

Simplify.

**endling** · October 22nd 2009, 08:38 PM

A run through:

Let's say your f(x) = g(h(q(x)))
With the chain rule it would be g'(h(q(x))) times h'(q(x)) times q'(x)
Then it's like what scorpion posted,
$g'(h(q(x))) = \frac{1}{\sqrt{1-\tan^2(3x^2)}}$
remember derivative of arcsin x = $\frac{1}{\sqrt{1-x^2}}$
$h'(q(x))= \sec^2(3x^2)$
$q'(x)= 6x$

**DontDrinkAndDerive** · October 22nd 2009, 08:50 PM

Thanks for the answer. I get how that goes about now.
Thanks Again

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