Fundamental Theorem of Calculus: Derivative of integral with respect to parameter in integration limits I have the following function:
$$F(t) = \int_{a-t}^{b+t} f(x) dx$$
In my notes when this is differentiated with respect to $t$ we get:
$$\frac{d F}{d t} = f(b + t) + f(a - t)$$
How can we obtain this result?
My approach
The approach that I was thinking about was about splitting the integral into two based on the integration limits, for instance, something like:
$$F_1(t) = \int_{a-t}^c f(x) dx \quad \quad F_2(t) = \int_{c}^{b+t} f(x)dx$$
So that $F(t) = F_1(t) + F_2(t)$, however I don't know how else to proceed.
 A: You're on the right track. You can proceed by using the chain rule.
Observe that (with your notation)
$$ F_1(t) = G(H(t)), $$
where
$$ H(t) = a-t, \quad \text{and} \quad G(t) = \int_t^c f(x) \, \mathrm dx = -\int_c^t f(x) \, \mathrm dx. $$
Now
$$ H'(t) = -1, \quad \text{and} \quad G'(t) = -f(t), $$
where we invoked FTC when differentiating $G$.
Hence, the chain rule gives
$$ F_1'(t) = G'(H(t))H'(t) = -G'(a-t)f(t) = f(a-t). $$
Can you deal with $F_2$ on your own?
P.S. You could also directly refer to the more general Leibniz integral rule.
A: If we have a function defined by $$Q'(x)=f(x)$$
Then we have that $$F(t)=Q(b+t)-Q(a-t)$$
Taking $\frac{\mathrm d}{\mathrm dt}(\cdot)$ on both sides:
$$F'(t)=\frac{\mathrm d}{\mathrm dt}Q(b+t)-\frac{\mathrm d}{\mathrm dt}Q(a-t)$$
$$F'(t)=Q'(b+t)\frac{\mathrm d}{\mathrm dt}(b+t)-Q'(a-t)\frac{\mathrm d}{\mathrm dt}(a-t)$$
$$F'(t)=Q'(b+t)+Q'(a-t)$$
$$F'(t)=f(b+t)+f(a-t)$$
QED
A: $$\frac{dF}{dt}=\frac{d(b+t)}{dt}\frac{d}{d(b+t)}\int_c^{b+t}f(x)dx-\frac{d(a-t)}{dt}\frac{d}{d(a-t)}\int_c^{a-t}f(x)dx\\=1\cdot f(b+t)-(-1)\cdot f(a-t)=f(b+t)+f(a-t).$$We use the chain rule at the first $=$, then the FTC at the second $=$.
