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).`
Now using the chain rule, we see that
since by the Fundamental Theorem, `F'(g(x))=f(g(x)).` This completes the proof.
If in the integral
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
`d/(dx)-` partial derivative w.r.t. x
`d/(dx)f(t)=0 , d/(dx)(1)=0 ,`
`` and `d/(dx)g(x)=g'(x)`