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(calculus) How can I prove that $$\frac{d}{dx}\frac{1}{\sqrt{1+x^2-\cos^2x-e^{2\pi \cos(\sin 1/x)}}}=\frac{-\frac{\displaystyle\pi\sin(\sin(1/x))\cos(1/x)e^{2\pi\cos(\sin(1/x))}}{x^2}+x+\sin x+\cos x}{(x^2-\cos^2x-e^{2\pi\cos(\sin(1/x))}+1)^{3/2}}$$

I have no idea how to start with, I've first simplified what's inside the square root to $\sin^2x+x^2-e^{2\pi\cos(\sin(1/x))}$ but then how to eradicate that square root? That's why I thought of using the chain rule where one of the functions is $1/\sqrt{x}$ but I'm having serious trouble!

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This is some brutal and brainless application of chain rule. Recall that $$\dfrac{d}{dx}\left(\dfrac1{\sqrt{g(x)}}\right) = -\dfrac12 \dfrac1{g(x)^{3/2}} \dfrac{dg(x)}{dx}$$ In your case, $$g(x) = 1+x^2 - \cos^2(x) - e^{2\pi \cos(\sin(1/x))}$$ and $$g'(x) = 2x + \sin(2x) - e^{2\pi \cos(\sin(1/x))} \cdot \left(2\pi \dfrac{d \cos(\sin(1/x))}{dx}\right)$$ where $$\dfrac{d \cos(\sin(1/x))}{dx} = -\sin\left(\sin(1/x)\right) \cos(1/x) \dfrac{-1}{x^2}$$ Put all these together and you will get what you want.

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I stress before that you work through this incase I've made an algebra slip

but anyway this is going to be a messy one, here goes;

so by the chain rule

$\frac{d}{dx} f(g(x)) = f'(g(x)).g'(x)$

and using where you got up to

$g(x) = 1+x^2 +sin^2(x)-e^{2\pi cos(sin(1/x))}$

and if we let $g(x) = y$

$f(x) = y^{-\frac{1}{2}}$

even in differentiating $g(x)$ we have a few messy bits so we do these seperately, lots of the working out is left out to make this as concise as possible;

$\frac {d}{dx} sin^2(x) = 2cos(x)sin(x)$

and by applying the chain rule multiple times on the $e^{...}$ term

$\frac {d}{dx} e^{2\pi cos(sin(1/x))} = \frac{2\pi}{x^2}sin(sin(1/x))cos(1/x)e^{2\pi cos(sin(1/x))} $

$g'(x) = 2x+2cos(x)sin(x)-\frac{2\pi}{x^2}sin(sin(1/x))cos(1/x)e^{2\pi cos(sin(1/x))}$

so now lets get $f'$

$\frac {d}{dx} f = -\frac{1}{2}y^{-\frac{3}{2}}$

so now by the chain rule we get

$\frac{d}{dx} f(g(x)) = f'(g(x))g'(x)$

$ = {(1+x^2 +sin^2(x)-e^{2\pi cos(sin(1/x))})}^{-\frac{3}{2}}(2x+2cos(x)sin(x)-\frac{2\pi}{x^2}sin(sin(1/x))cos(1/x)e^{2\pi cos(sin(1/x))}) $

which is equal to the expression you wanted

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