Overlapping Chain Rule?

**Biff** · November 13th 2012, 08:56 AM

Hi,

I'm having some trouble with this problem, I'm supposed to find the derivative:

$y = \sin^2(\cos(3x^{-2}))$

I know that the Chain Rule can be applied, but to what extent? There are three terms. I tried to simplify it to

$2\cos(\cos(3x^{-2}) \cdot -\sin(3x^{-2})$

Would the next step be to differentiate the right side? Also, on the left side, you could differentiate again and again only to still have the $\cos(3x^{-2})$ - how do you know for certain that once is enough?

Thanks in advance!

**Plato** · November 13th 2012, 09:21 AM

Originally Posted by Biff

find the derivative:
$y = \sin^2(\cos(3x^{-2}))$

Try
$\left\{ {2\sin \left( {\cos \left( {3x^{ - 2} } \right)} \right)} \right\}\left[ {\cos \left( {\cos \left( {3x^{ - 2} } \right)} \right)} \right]\left[ { - \sin \left( {3x^{ - 2} } \right)} \right]\left[ { - 6x^{ - 3} } \right]$

**HallsofIvy** · November 13th 2012, 09:46 AM

You can apply the chain rule as many times as you have strength for! Here, you have $y(x)= sin^2(cos(3x^{-2}))$
Let $u= 3x^{-2}$ . Then $y(u)= sin^2(cos(u))$ . Let v= cos(u). Then $y(v)= sin^2(v)$ . Let $w= sin(v)$ . Then $y(w)= w^2$ .
Then $\frac{dy}{dx}= \frac{dy}{dw}\frac{dw}{dv}\frac{dv}{du}\frac{du}{d x}$ which gives what Plato said.

**Biff** · November 13th 2012, 10:52 AM

Now this makes sense - thanks guys!

Here's how I solved it (separating it out helped):

$\frac{dy}{dx} = 2\sin(\cos(3x^{-2})) \cdot \cos(\cos(3x^{-2})) \cdot (-\sin(3x^{-2})) \cdot -6x^{-3}$

$\frac{dy}{dx} = 12x^{-3}\sin(\cos(3x^{-2})) \cdot \cos(\cos(3x^{-2})) \cdot \sin(3x^{-2}))$

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