If `y = (logcos x sinx)(logsin x cos x)–1 + sin–1(2x/(1 + x2))` then prove that `(dy/dx) at x = π/4 is 8(4/(π2 + 16) – 1/loge2)`

Explanation

We have `y=(log _{cos x} sin x)(log _{sin x} cos x)^{-1}+sin ^{-1}(2 x /(1+x^{2}))`


`=frac{log (sin x)}{log (cos x)} times frac{log (sin x)}{log (cos x)}+2 tan ^{-1} x`


`=frac{{log (sin x)}^{2}}{{log (cos x)}^{2}}+2 tan ^{-1} x`


`Rightarrow frac{d y}{d x}`


`{log (cos x)}^{2} 2{log (sin x)} cot x`


`=frac{+{log (sin x)}^{2} 2{log (cot x)} tan x}{{log (cos x)}^{4}}+frac{2}{1+x^{2}}`


`Rightarrow quad(frac{d y}{d x})_{x=frac{pi}{4}}`


`=frac{2{log (frac{1}{sqrt{2}})}^{3}+2{log (frac{1}{sqrt{2}})}^{3}}{{log (frac{1}{sqrt{2}})}^{4}}+frac{2}{1+(frac{pi}{4})^{2}}`


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