Why $\frac{d}{dy} \int_{-\infty}^{\frac{y-b}{a}}f(x)dx=\frac{1}{a}f \left ( \frac{y-b}{a} \right)$? Could somebody explain why:
$$\frac{d}{dy} \int_{-\infty}^{\frac{y-b}{a}}f(x)dx=\frac{1}{a}f \left ( \frac{y-b}{a} \right)$$
I don't get where this $\frac{1}{a}$ comes from. I assume that in general the outcome is based on the Leibnitz-Newton theorem:
$$\frac{d}{dy} \int_{a}^{y} f(x)dx$$
but I don't see how to get the proper result.
 A: Let $F(x)=\int f(x) dx$, 
so $\frac{d}{dy}\int_{-\infty}^{\frac{y-b}{a}}{f(x)}dx = \frac{d}{dy}[F(\frac{y-b}{a})-c] = \frac{1}{a}f(\frac{y-b}{a})$
A: We have to use two concepts:
1) Fundamental Theorem of Calculus
2) Chain rule of differentiation
Let $z = \dfrac{y - b}{a}$ then $\dfrac{dz}{dy} = \dfrac{1}{a}$. Next we have to calculate the value of $$\dfrac{d}{dy}\int_{-\infty}^{(y - b)/a}f(x)\,dx = \dfrac{d}{dy}\int_{-\infty}^{z}f(x)\,dx = \frac{d}{dy}F(z)$$ where we write $$F(z) = \int_{-\infty}^{z}f(x)\,dx$$
Now by fundamental theorem of calculus we get $\dfrac{d}{dz}\{F(z)\} = F'(z) = f(z)$ and by chain rule we get $$\frac{d}{dy}F(z) = \frac{d}{dz}\{F(z)\}\cdot \frac{dz}{dy} = f(z)\cdot\frac{1}{a} = \frac{1}{a}f\left(\frac{y - b}{a}\right)$$
Instead of taking into account multiple formulas and anti-derivatives it is better to stick to simple conceptual rules and theorems and the result is arrived quickly and simply.
A: In one of the books I found:
$$ \frac{d}{dy} \int_{a}^{\phi(y)} f(x) dx = f \left( \phi(y) \right) \cdot \phi '(y)$$
Which I understand better. Thanks a lot anyway :]
