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Question

If y=logcos x sin x logsin x cos x-1+sin-1 2x1+x2, find dydx at x=π4

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Solution

We have, y=logcosx sinxlogsinx cosx-1+sin-12x1+x2 y=logcosx sinxlogcosx sinx+sin-12x1+x2 logab=logba-1y=log sinxlog cosx2+sin-12x1+x2 logab=logbloga
Differentiating with respect to x,
dydx=ddxlog sinxlog cosx2+ddxsin-12x1+x2dydx=2log sinxlog cosxddxlog sinxlog cosx+11-2x1+x22×ddx2x1+x2 dydx=2log sinxlog cosxlog cosxddxlog sinx-log sinxddxlog cosxlog cosx2+1+x21+x4-2x21+x22-2x2x1+x22 dydx=2log sinxlog cosxlog cosx×1sinxddxsinx-log sinx×1cosxddxcosxlog cosx2+1+x21+x4-2x21+x22-2x2x1+x22 dydx=2log sinxlog cosxlog cosx×cosxsinx+log sinx × sinxcosxlog cosx2+1+x21-x222+2x2-4x21+x22dydx=2log sinxlog cosx3cotx log cosx+tanx log sinx+21+x2put x=π4dydx=2log sinπ4log cosπ43 cotπ4 log cosπ4+tanπ4 log sinπ4+211+π42dydx=21log1221× log12+1× log12+21616+π2 dydx=2×2log12log122+3216+π2dydx=41log12+3216+π2dydx=41-12log2+ 3216+π2dydx=-8log2+3216+π2So, dydxx=π4=8416+π2-1log2

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