How was this integral with functions as limits solved?

Question

How was this integral solved ?

$$\displaystyle \frac d{dy}\int_{g(y)}^{h(y)}f(x,y)dx$$ $$=\int_{g(y)}^{h(y)}\frac \partial{\partial y} f(x,y)dx+f(h(y),y)\frac{dh(y)}{dy}-f(g(y),y)\frac{dg(y)}{dy}$$

I can see intiutively where $+f(h(y),y)\frac{dh(y)}{dy}-f(g(y),y)\frac{dg(y)}{dy}$ come from but not the other term $\int_{g(y)}^{h(y)}\frac \partial{\partial y} f(x,y)dx$.

I do not see which was first calculated, the integral and the the derivative or was first the derivative and then the integral?

Google helps. It shows https://en.wikipedia.org/wiki/Leibniz_integral_rule, which has proofs. — MPW, May 21 '19 at 15:36
@MPW oh I didn't know it have a name, my book does not mention the name Leibniz — user441848, May 21 '19 at 15:41

score 1 · Accepted Answer · edited Jun 12 '20 at 10:38

$\displaystyle \frac d{dy}\int_{g(y)}^{h(y)}f(x,y)dx==\int_{g(y)}^{h(y)}\frac \partial{\partial y} f(x,y)dx+f(h(y),y)\frac{dh(y)}{dy}-f(g(y),y)\frac{dg(y)}{dy}$

The above expression is nothing but "Leibniz Integral Rule (Differentiation under the integral sign)".

Leibniz Integral Rule (Differentiation under the integral sign):

Let $f(x, t)$ be a function of $x$ and $t$ such that both $f(x, t)$ and its partial derivative $\frac{\partial f}{\partial x}$ are continuous in $t$ and $x$ in some region of the $(x, t)$-plane, including $a(x) ≤ t ≤ b(x)$, and $ x_0 ≤ x ≤ x_1$. Also suppose that the functions $a(x)$ and $b(x)$ are both continuous and both have continuous derivatives for $x_0 ≤ x ≤ x_1$. Then, for $x_0 ≤ x ≤ x_1$, $$\frac{d}{dx}\left(\int_{a(x)}^{b(x)} f(x,t) dt\right)=\int_{a(x)}^{b(x)} \frac{\partial }{\partial x}f(x,t) dt +f( x, b(x)) \frac{db}{dx}-f( x, a(x)) \frac{da}{dx}$$