Generally true that $\frac{\mathrm d}{\mathrm da} \int_{-\infty}^{a-y} f(x)\, \mathrm dx = f(a-y)$? Is it generally true that 
$$\frac{\mathrm d}{\mathrm da} \int_{-\infty}^{a-y} f(x)\, \mathrm dx = f(a-y),$$
where $a$ and $y$ in the expression  are constants?
To give context to the question, I am reading about the function convolution of two independent random variables $X, Y$. 
 A: Assuming $f$ is (for example) continuous, you are correct. To prove this, use the fundamental theorem of calculus in conjunction with the chain rule.
Let $$F(z)=\int_{-\infty}^{z} f(x)  \ dx.$$
By the fundamental theorem, $F'(z)=f(z)$. In your problem, you want to evaluate the derivative of $F(a-y)$ with respect to $a$. 
To use the chain rule, let's write $F(a-y)$ as a composition of two functions. Define $g(a)=a-y$. Then $F(a-y)=F(g(a))$. The derivative with respect to $a$, using the chain rule, is $F'(g(a))g'(a)$. We know $F'$ (it's $f$ -- see above), and we also know that $g'(a)=1$. 
So $F'(g(a))g'(a)=F'(g(a))=f(a-y)$. 
A: Here is a proof using the definition of derivative:
$$\begin{eqnarray*}\frac{d}{d a}\int_{\infty}^{a-y}f(x)dx &=& \lim_{t\to0}\frac{\int_{-\infty}^{a-y}f(x)dx-\int_{-\infty}^{a-y-t}f(x)dx}{t}\\
&=& \lim_{t\to0}\frac{\int_{a-y-t}^{a-y}f(x)dx}{t} \quad \textbf{mean value theorem for integration}\\
&=& \lim_{t\to0}f(\theta_t)\quad \textbf{where}\quad \theta_t\in(a-y-t,a-y)\\
&=& f(a-y)\end{eqnarray*}$$
