This is a question dealing with finding the derivative of a function

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Homework Statement



The problem is : f(x)= √3 cos^5(sinx²-3/³√x)

Homework Equations



The only thing i kno is that f prime of x comes next and that's all.. calculus is kind of confusing but i'll make sure i understand with your help. Thanks in advance

The Attempt at a Solution



S.O.S
 
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I don't understand finding the derivative of a functionn is hard please explan to me please. I really would appreciate it.
 
Is this the function you want to differentiate?
f(x)~=~\sqrt{3}cos^5\left(sin(x^2) - \frac{3}{\sqrt[3]{x}}\right)

Or is it this?
f(x)~=~\sqrt{3}cos^5\left(sin^2(x) - \frac{3}{\sqrt[3]{x}}\right)

Or something else?

Do you know the chain rule?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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