Find the derivative with respect to $x$ of the function:
$({log}_{cos x} sin x) ({log}_{sin x} cos x)^{- 1} + {sin}^{- 1} \frac{2 x}{1 + x^{2}}$ at $x = \frac{π}{4}$ .

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Solution

$y = ({log}_{cos x} sin x) ({log}_{sin x} cos x)^{- 1} + {sin}^{- 1} \frac{2 x}{1 + x^{2}} = (\frac{log sin x}{log cos x})^{2} + 2 {tan}^{- 1} x$
$\frac{d y}{d x} = 2 (\frac{log sin x}{log cos x}) (\frac{cot x log cos x + tan x log sin x}{(log cos x)^{2}}) + 2 \frac{1}{1 + x^{2}}$
At $x = \frac{π}{4}, \frac{d y}{d x} = 2 \times \frac{- log 2}{\frac{1}{4} (log 2)^{2}} + \frac{32}{π^{2} + 16} = - \frac{8}{log 2} + \frac{32}{π^{2} + 16}$

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({log}_{cos x} sin x) ({log}_{sin x} cos x)^{- 1} + {sin}^{- 1} \frac{2 x}{1 + x^{2}} at x = \frac{π}{4}

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