Chain Rule of a functional to an exponential

BreathingGloom
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Homework Statement


Suppose f is differentiable on \mathbb R and \alpha is a real number. Let G(x) = [f(x)]^a

Find the expression for G'(x)


Homework Equations



I'm pretty sure that I got this one right, but I really want to double check and make sure.

The Attempt at a Solution



G'(x) = a[f(x)]^{a-1} \cdot f'(x)
 
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Functional to an exponential.. haha. Not exactly what I meant, but okay.
 
That's correct. If you are being precise α ≠ 0.
 
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Much appreciated!
 

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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