Continuum mechanics/Leibniz formula

The Leibniz rule The integral $F(t)=\underset{a(t)}{\overset{b(t)}{\int }}f(x,t)\text{dx}$ is a function of the parameter $t$. Show that the derivative of $F$ is given by $\frac{dF}{dt}=\frac{d}{dt}(\underset{a(t)}{\overset{b(t)}{\int }}f(x,t)\text{dx})=\underset{a(t)}{\overset{b(t)}{\int }}\frac{\partial f(x,t)}{\partial t}\text{dx}+f[b(t),t]\frac{\partial b(t)}{\partial t}-f[a(t),t]\frac{\partial a(t)}{\partial t}.$ This relation is also known as the Leibniz rule.

Proof:

We have,

$\frac{dF}{dt}=\underset{\mathrm{\Delta }t\to 0}{\lim }\frac{F(t+\mathrm{\Delta }t)-F(t)}{\mathrm{\Delta }t}.$

Now,

$\begin{array}{cc}\frac{F(t+\mathrm{\Delta }t)-F(t)}{\mathrm{\Delta }t}& =\frac{1}{\mathrm{\Delta }t}[\underset{a(t+\mathrm{\Delta }t)}{\overset{b(t+\mathrm{\Delta }t)}{\int }}f(x,t+\mathrm{\Delta }t)\text{dx}-\underset{a(t)}{\overset{b(t)}{\int }}f(x,t)\text{dx}]\\ & \equiv \frac{1}{\mathrm{\Delta }t}[\underset{a+\mathrm{\Delta }a}{\overset{b+\mathrm{\Delta }b}{\int }}f(x,t+\mathrm{\Delta }t)\text{dx}-\underset{a}{\overset{b}{\int }}f(x,t)\text{dx}]\\ & =\frac{1}{\mathrm{\Delta }t}[-\underset{a}{\overset{a+\mathrm{\Delta }a}{\int }}f(x,t+\mathrm{\Delta }t)\text{dx}+\underset{a}{\overset{b+\mathrm{\Delta }b}{\int }}f(x,t+\mathrm{\Delta }t)\text{dx}-\underset{a}{\overset{b}{\int }}f(x,t)\text{dx}]\\ & =\frac{1}{\mathrm{\Delta }t}[-\underset{a}{\overset{a+\mathrm{\Delta }a}{\int }}f(x,t+\mathrm{\Delta }t)\text{dx}+\underset{a}{\overset{b}{\int }}f(x,t+\mathrm{\Delta }t)\text{dx}+\underset{b}{\overset{b+\mathrm{\Delta }b}{\int }}f(x,t+\mathrm{\Delta }t)\text{dx}-\underset{a}{\overset{b}{\int }}f(x,t)\text{dx}]\\ & =\underset{a}{\overset{b}{\int }}\frac{f(x,t+\mathrm{\Delta }t)-f(x,t)}{\mathrm{\Delta }t}\text{dx}+\frac{1}{\mathrm{\Delta }t}\underset{b}{\overset{b+\mathrm{\Delta }b}{\int }}f(x,t+\mathrm{\Delta }t)\text{dx}-\frac{1}{\mathrm{\Delta }t}\underset{a}{\overset{a+\mathrm{\Delta }a}{\int }}f(x,t+\mathrm{\Delta }t)\text{dx}.\end{array}$

Since $f(x,t)$ is essentially constant over the infinitesimal intervals $a<x<a+\mathrm{\Delta }a$ and $b<x<b+\mathrm{\Delta }b$, we may write

$\frac{F(t+\mathrm{\Delta }t)-F(t)}{\mathrm{\Delta }t}\approx \underset{a}{\overset{b}{\int }}\frac{f(x,t+\mathrm{\Delta }t)-f(x,t)}{\mathrm{\Delta }t}\text{dx}+f(b,t+\mathrm{\Delta }t)\frac{\mathrm{\Delta }b}{\mathrm{\Delta }t}-f(a,t+\mathrm{\Delta }t)\frac{\mathrm{\Delta }a}{\mathrm{\Delta }t}.$

Taking the limit as $\mathrm{\Delta }t\to 0$, we get

$\underset{\mathrm{\Delta }t\to 0}{\lim }\left[\frac{F(t+\mathrm{\Delta }t)-F(t)}{\mathrm{\Delta }t}\right]=\underset{\mathrm{\Delta }t\to 0}{\lim }\left[\underset{a}{\overset{b}{\int }}\frac{f(x,t+\mathrm{\Delta }t)-f(x,t)}{\mathrm{\Delta }t}\text{dx}\right]+\underset{\mathrm{\Delta }t\to 0}{\lim }[f(b,t+\mathrm{\Delta }t)\frac{\mathrm{\Delta }b}{\mathrm{\Delta }t}]-\underset{\mathrm{\Delta }t\to 0}{\lim }[f(a,t+\mathrm{\Delta }t)\frac{\mathrm{\Delta }a}{\mathrm{\Delta }t}]$

or,

$\frac{dF(t)}{dt}=\underset{a(t)}{\overset{b(t)}{\int }}\frac{\partial f(x,t)}{\partial t}\text{dx}+f[b(t),t]\frac{\partial b(t)}{\partial t}-f[a(t),t]\frac{\partial a(t)}{\partial t}.◻$

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