Why $\frac{d}{dy} \int_{-\infty}^{\frac{y-b}{a}}f(x)dx=\frac{1}{a}f \left ( \frac{y-b}{a} \right)$?

Question

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.

Answer 1

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.

Answer 2

Let $F(x)=\int f(x) dx$, so $\frac{d}{dy}\int_{-\infty}^{\frac{y-b}{a}}{f(x)}dx = \frac{d}{dy}[F(\frac{y-b}{a})-c] = \frac{1}{a}f(\frac{y-b}{a})$

Answer 3

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 :]