I'm probably overlooking something very simple, but I don't see what it is.
I want to determine the partial derivatives of the integral below, and I know the answers should be:

$\frac{d}{dx}(\int_0^x f(yt) \,dt)=f(yx)$
$\frac{d}{dy}(\int_0^x f(yt) \,dt)=f(yx)x$

But when I tried to write out why this was the case again, I got a different answer for the partial derivative with respect to x:

$\frac{d}{dx}(\int_0^x f(yt) \,dt)=\frac{d}{dx}(F(yx)-F(0))=f(yx)y$
$\frac{d}{dy}(\int_0^x f(yt) \,dt)=\frac{d}{dy}(F(yx)-F(0))=f(yx)x$
(Here $F$ denotes the antiderivative of $f$).

What am I doing wrong?


It should be: $$ \frac{\mathrm d}{\mathrm dx} \int_0^x f(yt) \, \mathrm dt = \frac{\mathrm d}{\mathrm dx} \frac{1}{y}F(yx) = f(yx) $$

  • $\begingroup$ Where does the 1/y term come from? $\endgroup$
    – A.Kawoela
    Jan 12 '20 at 13:27
  • 1
    $\begingroup$ Try doing it the opposite way: $\frac{\mathrm d}{\mathrm dx} F(yx) = yf(yx)$, so naturally for the reverse we need to include a factor of $\frac{1}{y}$. $\endgroup$ Jan 12 '20 at 13:29
  • $\begingroup$ ah I see, that makes sense. Thank you! $\endgroup$
    – A.Kawoela
    Jan 12 '20 at 13:36

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