d/dx integral (f^-1 (t) dt) the boundaries are 0 to f(x).

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- Dec 4th 2006, 01:47 PMmyoplex11FTOcalculus simplification problem
d/dx integral (f^-1 (t) dt) the boundaries are 0 to f(x).

- Dec 4th 2006, 04:31 PMThePerfectHacker
What makes so angry these problem do not mention intervals of a bijection map, continuity, definess, differenciability,.... Makes me so angry. I JUST CANNOT DO PROBLEMS LIKE THIS. So I am going to assume you have a differenciable and invertible function over the entire number line.

I presume you mean,

$\displaystyle g(x)=\int_0^{f(x)} f^{-1}(t)dt$

Then you can view $\displaystyle g$ as a composition,

$\displaystyle h\circ f$

Where,

$\displaystyle h=\int_0^x f^{-1}(t)dt$

Thus,

$\displaystyle h$ is differenciable.

Thus we can use chain rule theorem,

$\displaystyle g'=f'\cdot (h'\circ f)$

But by fundamental theorem,

$\displaystyle h'=f(x)$

Thus,

$\displaystyle g'=f'\cdot (f\circ f^{-1})$

But,

$\displaystyle f\circ f^{-1}=x$ the identity map.

Thus,

$\displaystyle g'=xf'$