# Partial derivatives of definite integral

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?

## 1 Answer

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

• Where does the 1/y term come from? Jan 12 '20 at 13:27
• 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 $\frac{1}{y}$. Jan 12 '20 at 13:29
• ah I see, that makes sense. Thank you! Jan 12 '20 at 13:36