How can I derive this formula using chain rule?

Question

Let $f:\mathbb{R}^2 \to \mathbb{R}$ be a continuously differentiable function which vanishes outside a bounded set in $\mathbb{R}^2$ and let $h:\mathbb{R} \to \mathbb{R}$ be a continuously differentiable function. The given result is
$$D\left(\int_{-\infty}^{h(x)}f(x,t)dt\right) =f(x,h(x))h'(x) + \int_{-\infty}^{h(x)}D_1f(x,t)dt $$

I cannot capture how I could differentiate the given function using chain rule.

Answer 1

First note that since $f(x,t) = 0$ for all $(x,t)$ outside of some bounded set, we know there exists some real number $M$ such that $f(x,t) = 0$ for all $t<M$. Therefore $$ \int_{-\infty}^{h(x)} f(x,t) \,dt = \int_M^{h(x)} f(x,t) \,dt $$

Next, let the function $F$ denote a partial antiderivative of $f$ with respect to $t$. That is, $F$ satisfies \begin{align} \frac{\partial F}{\partial t} = f(x,t) \end{align} Then$\newcommand{\ddx}{\frac{d}{dx}}$ \begin{align} \ddx \int_{M}^{h(x)} f(x,t) \,dt &= \ddx \big[F(x,t) \big]_{t=M}^{h(x)} \\ &= \ddx \big[F(x,h(x)) - F(x,M) \big] \end{align}

At this point we need the multivariable chain rule. I don't know how familiar you are with it, but the version of it we need is this:

Consider a function $f(x,y)$ where both $x$ and $y$ are themselves functions of a variable $t$. Then $$ \frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$$

Therefore we get \begin{align} \underset{{(x,h(x))}}{\frac{\partial F}{\partial x}} \frac{dx}{dx} + \underset{(x,h(x))}{\frac{\partial F}{\partial t}} \frac{dh}{dx} - \left[\underset{(x,M)}{\frac{\partial F}{\partial x}} \frac{dx}{dx} + \underset{(x,M)}{\frac{\partial F}{\partial t}} \frac{dM}{dx} \right] \end{align} where the notation $$\underset{(a,b)}{\frac{\partial F}{\partial x}}$$ means $\partial F/\partial x$ evaluated for $x=a$ and $t=b$.

Now, $dx/dx = 1$ of course, and $M$ is a constant, so $dM/dx = 0$. Also $\partial F/\partial t = f(x,t)$ by definition of $F$. So our last expression becomes \begin{align} &= \underset{(x,h(x))}{\frac{\partial F}{\partial x}} + f(x,h(x))\,h'(x) - \underset{(x,M)}{\frac{\partial F}{\partial x}} \\ &= f(x,h(x))\,h'(x) + \left[\underset{(x,h(x))}{\frac{\partial F}{\partial x}} - \underset{(x,M)}{\frac{\partial F}{\partial x}} \right] \ \ \ \ \textrm{Rearranging terms} \\ &= f(x,h(x))\,h'(x) + \int_M^{h(x)} \frac{\partial}{\partial t} \frac{\partial F}{\partial x}\,dt \ \ \ \ \textrm{Fundamental Theorem of Calculus} \\ &= f(x,h(x))\,h'(x) + \int_M^{h(x)} \frac{\partial}{\partial x} \frac{\partial F}{\partial t}\,dt \ \ \ \ \textrm{Clairaut's Theorem} \\ &= f(x,h(x))\,h'(x) + \int_M^{h(x)} \frac{\partial}{\partial x} f(x,t)\,dt \ \ \ \ \textrm{Definition of $F$} \\ &= f(x,h(x))\,h'(x) + \int_{-\infty}^{h(x)} \frac{\partial}{\partial x} f(x,t)\,dt \ \ \ \ \textrm{because $f = 0$ for all $t<M$} \end{align}