Leibnitz’s Rule
Leibnitz’s Rule: If f is a continuous function on [a, b], and u(x) and v(x) are differentiable function of x whose values lie in [a, b], then \(\frac{d}{dx}\left\{ \int\limits_{u(x)}^{v(x)}{f(t).dt} \right\}=f\left( v\left( x \right) \right)\frac{dv\left( x \right)}{dx}-f\left( u\left( x \right) \right)\frac{du\left( x \right)}{dx}\).
Proof:
Let d/dx(F(x)) = f(x)
\(\int\limits_{u(x)}^{v(x)}{f(t)dt}=F\left( v(x) \right)-F\left( u(x) \right)\),
\(\frac{d}{dx}\left\{ \int\limits_{u(x)}^{v(x)}{f(t)dt} \right\}=\frac{d}{dx}\left( F\left( v(x) \right)-F\left( u(x) \right) \right)\),
\(\frac{d}{dx}\left\{ \int\limits_{u(x)}^{v(x)}{f(t)dt} \right\}=F’\left( v(x) \right)\frac{d(v(x))}{dx}-F’\left( u(x) \right)\frac{d(u(x))}{dx}\),
\(\frac{d}{dx}\left\{ \int\limits_{u(x)}^{v(x)}{f(t)dt} \right\}=f\left( v(x) \right)\frac{d(v(x))}{dx}-f\left( u(x) \right)\frac{d(u(x))}{dx}\).
Example: If \(y=\int\limits_{{{x}^{2}}}^{{{x}^{3}}}{\frac{1}{\log t}}.dt\) (where x > 0), then find dy/dx.
Solution:
\(y=\int\limits_{{{x}^{2}}}^{{{x}^{3}}}{\frac{1}{\log t}}.dt\),
\(\frac{dy}{dx}=\frac{d}{dx}\left( {{x}^{3}} \right)\frac{1}{\log {{x}^{3}}}-\frac{d}{dx}\left( {{x}^{2}} \right)\frac{1}{\log {{x}^{2}}}\),
\(\frac{dy}{dx}=\left( 3{{x}^{2}} \right)\frac{1}{\log {{x}^{3}}}-\left( 2x \right)\frac{1}{\log {{x}^{2}}}\),
\(\frac{dy}{dx}=\left( 3{{x}^{2}} \right)\frac{1}{3\log x}-\left( 2x \right)\frac{1}{2\log x}\)\(\frac{dy}{dx}=\left( 3{{x}^{2}} \right)\frac{1}{3\log x}-\left( 2x \right)\frac{1}{2\log x}\),
\(\frac{dy}{dx}=\frac{3{{x}^{2}}}{3\log x}-\frac{2x}{2\log x}\),
\(\frac{dy}{dx}=\frac{{{x}^{2}}}{\log x}-\frac{x}{\log x}\),
\(\frac{dy}{dx}={{x}^{2}}\log {{x}^{-1}}-x\log {{x}^{-1}}\),
\(\frac{dy}{dx}=\log {{x}^{-1}}\left( {{x}^{2}}-x \right)\),
\(\frac{dy}{dx}=\log {{x}^{-1}}x\left( x-1 \right)\).