How to differentiate $f(a(y),y)$ with respect to $y$? What rule do we need to apply on it?
Actually, I'm trying to understand the Leibniz’s Rule. $$\displaystyle\frac{d}{d_y}\int_{a(y)}^{b(y)}f(x,y)dx=f(b(y),y)·\frac{d}{d_y}b(y)-f(a(y),y)·\frac{d}{d_y}a(y)+\int_{a(y)}^{b(y)}\frac{\partial}{\partial y}f(x,y)dx$$
What I tried to do was
Let $\displaystyle G(y)=\int_{a(y)}^{b(y)}f(x,y)dx=F(b(y),y)-F(a(y),y)$
$$\displaystyle\frac{d}{d_y}\int_{a(y)}^{b(y)}f(x,y)dx=\frac{d}{d_y}G(y)=\frac{d}{d_y}[F(b(y),y)-F(a(y),y)]=f(b(y),y)·\frac{d}{d_y}b(y)-f(a(y),y)·\frac{d}{d_y}a(y)$$
But obviously, I lost the last term $\int_{a(y)}^{b(y)}\frac{\partial}{\partial y}f(x,y)dx$
You have: $$\frac{df(a,y)}{dy}=\frac{\partial f}{\partial a}\frac{da}{dy}+\frac{\partial f}{\partial y}=\left(a'\partial_a+\partial_y\right)f(a(y),y)$$ However this is not the best way to work out a proof for the Liebniz integral rule. Use the fact that: $$\frac{df(x)}{dx}=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$ so we can do this to our integral: $$I'=\frac{d}{dy}\int_{a(y)}^{b(y)}f(x,y)dx=\lim_{h\to0}\frac 1h\left[\int_{a(y+h)}^{b(y+h)}f(x,y+h)dx-\int_{a(y)}^{b(y)}f(x,y)dy\right]$$ Now we can manipulate this in some ways by noticing: $$\int_{a(y+h)}^{b(y+h)}f(x,y+h)dx=\int_{a(y)}^{b(y)}f(x,y+h)dx+\int_{b(y)}^{b(y+h)}f(x,y+h)dx-\int_{a(y)}^{a(y+h)}f(x,y+h)dx$$
Now the terms in our limit become: $$\int_{a(y)}^{b(y)}\frac{f(x,y+h)-f(x,y)}{h}dx+\frac 1h\int_{b(y)}^{b(y+h)}f(x,y+h)dx-\frac 1h\int_{a(y)}^{a(y+h)}f(x,y+h)dx$$ you should be able to see now where this is going
$$I'=\int_{a(y)}^{b(y)}\partial_yf(x,y)dx+b'f(b,y)-a'f(a,y)$$