How to get `w'(x) = f(g(x)) g'(x)` from `w(x) = int_1^g(x)f(t)dt`

2 Answers

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degeneratecircle | High School Teacher | (Level 2) Associate Educator

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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


Now using the chain rule, we see that


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

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pramodpandey | College Teacher | (Level 3) Valedictorian

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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


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

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

`` and   `d/(dx)g(x)=g'(x)`