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I was doing some elementary derivatives when I came across the following. $$f(x)=\sqrt{\frac{1-x}{1+x}}\frac{\sin^3x}{\cos x}$$ Find $f'(x).$

I can see that product rule and quotient rule can give an answer but the solution is going to be a bit long. So, my question is that can I solve this problem in a 'simpler' and 'faster' way. However, all kinds of solutions and suggestions are welcome.

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Take the $\log$ of both sides.

You then get $\displaystyle \log(f(x))=\log\left(\sqrt{\frac{1-x}{1+x}}\frac{\sin^3x}{\cos x}\right)$.

Expand the logarithm to get $\displaystyle \log(f(x))=\frac{1}{2}\log(1-x)-\frac{1}{2}\log(1+x)+3\log(\sin(x))-\log(\cos(x))$.

Now differentiate (and use chain rule where necessary) to get $\displaystyle \frac{f'(x)}{f(x)}=\frac{1}{2(x-1)}-\frac{1}{2(1+x)}+3\cot(x)+\tan(x).$

Then multiply by $f(x)$ on both sides to get

$\displaystyle \boxed{f'(x)=\left(\sqrt{\frac{1-x}{1+x}}\frac{\sin^3x}{\cos x}\right)\left(\frac{1}{2(x-1)}-\frac{1}{2(1+x)}+3\cot(x)+\tan(x)\right)}$

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Perhaps writing $$\frac{\sin^3 x}{\cos x}=\frac{\sin x\left(1-\cos^2 x\right)}{\cos x}=\frac{\sin x}{\cos x}-\frac{\sin x\cos^2 x}{\cos x}=\tan x-\frac{1}{2}\sin\left(2x\right)$$ will make it easier.

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you can also substitute $x=\cos(t)$ then we have $1-x,1+x$ as $2\sin^2(\frac{t}{2}),2\cos^2(\frac{t}{2})$. We can also simplify $\frac{sin^3(x)}{cos(x)}=\tan(x)-\frac{1}{2}\sin(2x)$ as pointed out by Guy. Then we can calculate $\frac{dx}{dt}=\frac{1}{\frac{dt}{dx}}$ and get the result.

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