Math Help - What is the derivative of an integral wrt the integrand?

1. What is the derivative of an integral wrt the integrand?

By the fundamental thm of calc, the derivative of a definite integral wrt to the upper limit is the integrand evaluated at the upper limit.

$\frac{d}{dy} \int^{y}_{0} f(x) dx = f(y)$

What if I integrate wrt to the integrand?

$\frac{d}{df(x)} \int^{y}_{0} f(x) dx = ?$

Thanks!

2. Originally Posted by richardcherron
By the fundamental thm of calc, the derivative of a definite integral wrt to the upper limit is the integrand evaluated at the upper limit.

$\frac{d}{dy} \int^{y}_{0} f(x) dx = f(y)$

What if I integrate wrt to the integrand?

$\frac{d}{df(x)} \int^{y}_{0} f(x) dx = ?$

Thanks!
$\frac{d}{dy}\int_0^y{f}(x)dx$

$=\frac{d}{dy}F(x)\Big|_0^y$

$=\frac{d}{dy}[F(y)-F(0)]$. But note that $F(0)$ is constant which implies that

$\frac{d}{dy}[F(y)-F(0)]=\frac{d}{dy}[F(y)]+0=f(y)$

3. Oh, wait. I see what you're after.

$\frac{d}{df(x)}\int_0^y{f(x)dx}=\frac{d}{df(x)}[F(y)]$.

So, if $u=f(x)$, and $v=F(y)$ we have

$\frac{dv}{du}$

4. Originally Posted by VonNemo19
$\frac{d}{dy}\int_0^y{f}(x)dx$

$=\frac{d}{dy}F(x)\Big|_0^y$

$=\frac{d}{dy}[F(y)-F(0)]$. But note that $F(0)$ is constant which implies that

$\frac{d}{dy}[F(y)-F(0)]=\frac{d}{dy}[F(y)]+0=f(y)$
I guess you misread, since he wants the derivative wrt f.

Anyway, maybe you can write : $\frac{d}{df({\color{red}y})}\int_0^y f(x) ~dx=\frac{dy}{df(y)} \cdot \frac{d}{dy} \int_0^y f(x) ~dx$

$=\frac{1}{f'(y)} \cdot f(y)=\frac{f(y)}{f'(y)}$