relation

f(x)=ln (1+sin ^2x )

what is g'(x)?

g(x)=integral S (pi /4 down _ asin x up ) f(t)dt

### 1 Answer | Add Yours

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)"`

Ans.

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes