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Asked by | 16 May, 2010, 12:01: AM Expert Answer

logcosxsinx = 1/logsinxcosx, hence,

y = {logcosxsinx }2 + sin-1{2x/(1+x2)}

Using change of base,

y = {ln sinx / ln cosx }2 + sin-1{2x/(1+x2)}

y = F(x) + G(x),

dy/dx = dF(x)/dx + dG(x)/dx;

dF(x)/dx = (2 ln sinx / ln cosx)(ln cosx . 1/sinx . cosx  - ln sinx. 1/cosx . -sinx)/(ln cosx) 2

At x = π/4, sin x = cos x = 1/ 2,

Hence dF(x)/dx at  x = π/4, = 4 ln (1/ 2) / (ln (1/ 2))2 = 4 ln 2 = ln 4

Now, dG(x)/dx = 1/ (1 - {2x/(1+x2)}2) . [{(1+x2).2 - 2x(2x)}/(1+x2)2]

dG(x)/dx = 1/ (1 - {2x/(1+x2)}2) . 2[(1 - x2)/(1+x2)2]

dG(x)/dx = (1+x2)/ ( (1+x2)2 - 4x2) . 2[(1 - x2)/(1+x2)2]

dG(x)/dx = 1/ ( (1+x2)2 - 4x2) . 2[(1 - x2)/(1+x2)]

dG(x)/dx = 1/ (1- x2)  . 2[(1 - x2)/(1+x2)]

dG(x)/dx = 2/(1+x2)

dG(x)/dx, at x =  = π/4, = 2/(1 + π2/16)

dG(x)/dx, at x =  = π/4, = 32/(16 + π2)

Hence dy/dx = dF(x)/dx + dG(x)/dx = ln 4  + 32/(16 + π2)

Regards,

Team,

TopperLearning.

Answered by | 19 May, 2010, 12:06: PM

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