How to differentiate definite integral ?

Hint: Definite integral are those integration which have limits, for example $\int\limits_{a}^{b}{f\left( x \right)}dx$ is a definite integral first we will integrate function f , If the limits of the integral is constant number then the derivative value is 0. If the limits are functions of some variable , we can find the derivative by Leibniz rule. Leibniz rule states $\dfrac{d}{dx}\int\limits_{b\left( x \right)}^{a\left( x \right)}{f\left( x,t \right)dt}=\int\limits_{b\left( x \right)}^{a\left( x \right)}{\left( \dfrac{\partial }{\partial x}f\left( x,t \right) \right)}dt+f\left( x,a\left( x \right) \right)\dfrac{\partial a\left( x \right)}{\partial x}-f\left( x,b\left( x \right) \right)\dfrac{\partial b\left( x \right)}{\partial x}$

Complete step-by-step solution:
When we differentiate a integral, if the limits are function of some variable, for example if we have to differentiate $\int\limits_{b\left( t \right)}^{a\left( t \right)}{f\left( x,t \right)}dt$ with respect to x. First we can differentiate the function f with respect to x and then integrate with respect to t. then we have to add the result with $f\left( x,a\left( x \right) \right)\dfrac{\partial a\left( x \right)}{\partial x}-f\left( x,b\left( x \right) \right)\dfrac{\partial b\left( x \right)}{\partial x}$ .
If the function f is pure function of t, then $\left( \dfrac{\partial }{\partial x}f\left( t \right) \right)$ will be equal to 0, so the result will be $f\left( a\left( x \right) \right)\dfrac{\partial a\left( x \right)}{\partial x}-f\left( b\left( x \right) \right)\dfrac{\partial b\left( x \right)}{\partial x}$

Note: We will solve a example by using above formula , let's take function f is equal to $\log \left( x+t \right)$ . let’s take the limits are form x to ${{x}^{2}}$ .The value of $\dfrac{d}{dx}\int\limits_{b\left( x \right)}^{a\left( x \right)}{\log \left( x+t \right)dt}$ is equal to $\int\limits_{x}^{{{x}^{2}}}{\left( \dfrac{\partial }{\partial x}\log \left( x+t \right) \right)}dt+\log \left( x+{{x}^{2}} \right)\dfrac{\partial {{x}^{2}}}{\partial x}-\log \left( x+x \right)\dfrac{\partial x}{\partial x}$ . We know that the derivative of ${{x}^{2}}$ with respect to x is equal to 2x and derivative of x is equal to 1. After solving it we get $\dfrac{d}{dx}\int\limits_{b\left( x \right)}^{a\left( x \right)}{\log \left( x+t \right)dt}$ is equal to $\log \left( x+{{x}^{2}} \right)-\log \left( 2x \right)-\log \left( x \right)+2x\log \left( x+{{x}^{2}} \right)-\log \left( 2x \right)$ . if we solve further, we get
$\Rightarrow \dfrac{d}{dx}\int\limits_{b\left( x \right)}^{a\left( x \right)}{\log \left( x+t \right)dt}=\left( 2x+1 \right)\log \left( x+{{x}^{2}} \right)-\log \left( 4{{x}^{3}} \right)$ . From this formula we can easily derive that the value of $\dfrac{d}{dx}\left( \int\limits_{c}^{x}{f\left( t \right)dt} \right)$ is equal to f ( x ) where c is any constant number.