A simple derivative that I'm messing up on

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Homework Statement



Find f'(x) for f(x) = sin x + 2 cos^3 x

Homework Equations



Other than basic derivative properties and formulas, no.

The Attempt at a Solution



f'(x) = sin x + 2cos^3 x
= cos x - 6sin^2 x

The book says the answer is f'(x) = cos x - 6cos^2 x sin x however I don't understand where the sin x comes from... any help?
 
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Do you know the chain rule?
 
Yeah, d/dx[f(g(x))] = f'(g(x))g'(x). My bad, now I got the correct answer of f'(x) = cos x - 6cos^2 x sin x.. I think I used the power rule on 2cos^3 x by mistake before.
 
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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