CBSE Class 12-science Answered
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)
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