relation  f(x)=ln (1+sin ^2x ) what  is  g'(x)? g(x)=integral  S (pi /4 down  _ asin x  up ) f(t)dt 

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We know Leibnitz rule for differentiation under integral sign.

Let differentiation under integral sign is valid.

`g(x)=int_(pi/4)^(asinx)ln(1+sin^2t)dt`

differentiate g(x) with respect to x,we have By Leibniz rule,

`g'(x)=int_(pi/4)^(asinx)(d(ln(1+sin^2t)))/dxdt+`

`ln(1+sin^2t)(d(asinx))/dx-ln(1+sin^2t)(d(pi/4))/dx`

`` `g'(x)=ln(1+sin^2t)(acosx)`

`(d(ln(1+sin^2t)))/dx=` partial derivative of ln(1+sin^2t) with respect to

x     = 0

`"g'(x)=ln(1+sin^2t)(acosx)"`

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