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How to find the derivative of $y =(\log_{\cos x}\sin x)(\log_{\sin x}\cos x)+\sin^{-1}\frac{2x}{1+x^2}$

Please guide...thanks..

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Hint: do you know of a way to express $\log_a b$ using only natural logarithms? –  Javier Badia May 2 '13 at 2:41
    
The first term is a distraction: what is the product $(\log_a b) \cdot (\log_b a)$ equal to? –  RecklessReckoner May 2 '13 at 2:42

2 Answers 2

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Use base change to get that: $$\log_a b\cdot \log_b a = \frac{\log_b b}{\log_b a}\cdot \log_b a =1$$ Then you're left with $$y =\sin^{-1}\frac{2x}{1+x^2}$$ The derivative of $\sin^{-1}$ is $(1-x^2)^{-1/2}$, so: $$y' =\left[1-\left(\frac{2x}{1+x^2}\right)^2\right]^{-1/2}\times \frac{2(1+x^2)-4x^2}{(1-x^2)^{2}}$$

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$$\text{For }\sin^{-1}\left(\frac{2x}{1+x^2}\right)$$

Putting $x=\tan\theta, \frac{2x}{1+x^2}=\frac{2\tan\theta}{1+\tan^2\theta}=2\sin\theta\cos\theta=\sin2\theta$

$\implies \sin^{-1}\left(\frac{2x}{1+x^2}\right)=2\theta=2\tan^{-1}x$

$$\text{Now, }\frac{d(\tan^{-1}x)}{dx}=\frac1{1+x^2}$$

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@sultan, there is a simpler way to differentiate $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ –  lab bhattacharjee May 2 '13 at 4:09

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