Problem in Tom M Apostol's Calculus Vol-II book regarding Partial Derivatives.

Question

I've found some difficulties in solving the following problem from Apostol, Calculus Vol-II: p-292

If $k$ is a positive constant and

$$g(x,t)=\frac{x}{2\sqrt{kt}}$$

let

$$f(x,t)= \int_{0}^{g(x,t)}e^{-u^2}du$$

Show that

$$\frac{\partial f}{\partial x}= e^{-g^2}\frac{\partial g}{\partial x}$$

and

$$\frac{\partial f}{\partial t}= e^{-g^2}\frac{\partial g}{\partial t}$$

In this problem I want to use only the definitions of partial derivatives and basic theorems on that (as demanded by the exercise).

My Solution

I try to find the partial derivative of $f$ with respect to $x$. So we have

\begin{align} D_{1}f(x,t) &= \lim_{h\to 0}\frac{f(x+h, t)-f(x,t)}{h} \\ &=\lim_{h\to 0}\frac{1}{h} \left[\int_{0}^{g(x+h,t)}e^{-u^2}du-\int_{0}^{g(x,t)}e^{-u^2}du \right] \\ &=\lim_{h\to 0}\frac{1}{h}\int_{g(x,t)}^{g(x+h,t)}e^{-u^2}du \quad \text{(since, for fixed $x$ and $ t$, $g(x,t)$ is increasing)} \end{align}

Here, I can't figure out this limit. Also, for the other one, we have

$$D_2f(x,t)=\lim_{h\to 0}\frac{1}{h}\int_{g(x,t+h)}^{g(x,t)}e^{-u^2}du$$

Please help.

Answer 1

Similar to Chris Custer's answer, consider the general case of $$f(x,t)=\int_0^{g(x,t)} h(u) \, du$$ Then, using the fundamental theorem of calculus, $$\frac{\partial f}{\partial x}= h(g(x,t))\, \frac{\partial g(x,t)}{\partial x}$$ $$\frac{\partial f}{\partial t}= h(g(x,t))\, \frac{\partial g(x,t)}{\partial t}$$ and we can continue for higher derivatives using the chain rule.

Answer 2

The two partials are computed by using the fundamental theorem of calculus (FTC), and the chain rule...

Since the upper limit of integration is a function of $x$ and $t$, you need the chain rule... this means multiplying by $\frac{\partial g}{\partial x}$ and $\frac{\partial g}{\partial t}$ respectively...

FTC just tells us to replace the variable ($u$) in the integrand with the upper limit of integration ($g$)...

The result follows immediately.