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I have tried the integral below with Mathematica and it gives me the following solution:

##\frac{d}{dc}\int_{z^{-1}(c)}^{1} z(x)dx = -\frac{c}{z'(z^{-1}(c))}##

I am not quite sure where it gets it from...I think it can be separated and with differentiation the first part will be zero:

##\frac{d}{dc}\int_{z^{-1}(c)}^{1}z(x)dx=\frac{d}{dc}\int_{0}^{1} z(x)dx-\frac{d}{dc}\int_{0}^{z^{-1}(c)} z(x)dx=-\frac{d}{dc}\int_{0}^{z^{-1}(c)} z(x)dx##

By the formula ##\frac{d}{dc}z^{-1}(c) = \frac{1}{z'(z^{-1}(c))}##, so it definitely has to play a role here, but how? and where does c come from? Do we need to somehow use chain rule here?

If the upper limit was just c it would be easy to take it with a Leibniz rule:

##-\frac{d}{dc}\int_{0}^{c} z(x)dx=z(c)##

However, it seems that when one of the limits is an inverse function, there is something different at work which I am not aware of...

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# A Integral with an inverse function limit

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