How do I use the chain and product rules together?

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Ok, I have having a very hard time finding the derivative of this, I have no clue how to do the set up.

x^2(sqrt(9-x^2))

(sorry it's written like that I just have no clue how to type it normally)
So the problem I am having is When do I use the chain rule, and the product rule cohesively, I have the answer to the problem but it does me no good because I cannot solve it.
Thanks in advance.
 
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Any help would be greatly appreciated, I have been pulling my hair out for 2 hours :(
 
You have a product, so you need to use the product rule first. One of the factors is a composite function, so after you use the product rule, you'll need to use the chain rule.

Don't try to do everything at once.

d/dx[x^2 * (9 - x^2)^(1/2)] = x^2 * d/dx[(9 - x^2)^(1/2)] + 2x * (9 - x^2)^(1/2) = ?
 
Thank you so much that cleared it right up for me :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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