It is often considered part of the Fundamental Theorem of Calculus that if `f` is continuous on an interval containing `[1,x],` and if we define

`F(x)=int_1^x f(t) dt,` then `F'(x)=f(x).`

Note that

`w(x)=F(g(x))=int_1^g(x)f(t)dt.`

Now using the chain rule, we see that

`w'(x)=F'(g(x))g'(x)=f(g(x))g'(x),`

since by the Fundamental Theorem, `F'(g(x))=f(g(x)).` This completes the proof.

If in the integral

`w(x)=int_1^g(x)f(t)dt` (i)

f(t) is fuction of single variable i.e ' t '

Its ( partial )derivative w.r.t. x will zero.

Aply Leibnitz rule ,for differentiation ubder integral sign.

Let differentiation ubder integral sign is valid. So differentiate (i) with resp. to x ,we have

`w'(x)=int_1^g(x)(d/(dx)f(t))dt+f(g(x))d/(dx)g(x)-f(1)d/(dx)(1)`

`d/(dx)-` partial derivative w.r.t. x

`d/(dx)f(t)=0 , d/(dx)(1)=0 ,`

`` and `d/(dx)g(x)=g'(x)`

So

`w'(x)=f(g(x))g'(x)`

Ans.