For a function $f\left( x \right)$, if $\int\limits_{0}^{x}{f\left( t \right)dt={{x}^{2}}+\int\limits_{x}^{1}{{{t}^{2}}f\left( t \right)}}dt$, then $f'\left( \dfrac{1}{2} \right)$ is:
A. $\dfrac{6}{25}$
B. $\dfrac{24}{25}$
C. $\dfrac{18}{25}$
D. $\dfrac{4}{5}$

Hint: To solve this question, we should know the Leibnitz rule in calculus. Given the equation $\int\limits_{a\left( x \right)}^{b\left( x \right)}{f\left( t \right)dt=}g\left( x \right)$, the differentiation with respect to x gives
$\begin{align}
  & \dfrac{d}{dx}\left( \int\limits_{a\left( x \right)}^{b\left( x \right)}{f\left( t \right)dt} \right)=\dfrac{d}{dx}\left( g\left( x \right) \right) \\
 & f\left( b\left( x \right) \right)b'\left( x \right)-f\left( a\left( x \right) \right)a'\left( x \right)=g'\left( x \right) \\
\end{align}$
By using the above rule in the given equation $\int\limits_{0}^{x}{f\left( t \right)dt={{x}^{2}}+\int\limits_{x}^{1}{{{t}^{2}}f\left( t \right)}}dt$ and differentiating with respect to x, we get the result.

Complete step-by-step answer:
Let us consider the integral
$\int\limits_{a\left( x \right)}^{b\left( x \right)}{f\left( t \right)dt=}g\left( x \right)$
Let us consider that $\int{f\left( x \right)dx}=F\left( x \right)$
We can write the above integral as $\left[ F\left( x \right) \right]_{a\left( x \right)}^{b\left( x \right)}=g\left( x \right)$. By substituting the values of a, b, we get
$F\left( b\left( x \right) \right)-F\left( a\left( x \right) \right)=g\left( x \right)$.
Now, let us differentiate the above equation with respect to x, we get
$\dfrac{d}{dx}\left( F\left( b\left( x \right) \right)-F\left( a\left( x \right) \right) \right)=\dfrac{d}{dx}\left( g\left( x \right) \right)$
We know the chain rule in differentiation which is
$\dfrac{d}{dx}\left( F\left( b\left( x \right) \right) \right)=F'\left( b\left( x \right) \right)\times b'\left( x \right)$
Using this result in the above equation, we get
$F'\left( b\left( x \right) \right)b'\left( x \right)-F'\left( a\left( x \right) \right)a'\left( x \right)=g'\left( x \right)$
We know that $F'\left( x \right)=f\left( x \right)$, using it in above equation, we get
$f\left( b\left( x \right) \right)b'\left( x \right)-f\left( a\left( x \right) \right)a'\left( x \right)=g'\left( x \right)$
So, we can finally write that differentiating $\int\limits_{a\left( x \right)}^{b\left( x \right)}{f\left( t \right)dt=}g\left( x \right)$ gives
$f\left( b\left( x \right) \right)b'\left( x \right)-f\left( a\left( x \right) \right)a'\left( x \right)=g'\left( x \right)$.
This is called Leibnitz rule in calculus.
We are given the equation $\int\limits_{0}^{x}{f\left( t \right)dt={{x}^{2}}+\int\limits_{x}^{1}{{{t}^{2}}f\left( t \right)}}dt$.
Let us consider differentiating the equation with respect to x. We get
$f\left( x \right)\dfrac{d}{dx}\left( x \right)-f\left( 0 \right)\dfrac{d}{dx}\left( 0 \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right)+{{1}^{2}}\times f\left( 1 \right)\dfrac{d}{dx}\left( 1 \right)-{{x}^{2}}\times f\left( x \right)\dfrac{d}{dx}\left( x \right)$
We know that differentiating a constant will give zero as the value. Using it, we get
$f\left( x \right)\times 1-0=2x+0-{{x}^{2}}\times f\left( x \right)\times 1$
Further simplifying, we get
$\begin{align}
  & f\left( x \right)+{{x}^{2}}f\left( x \right)=2x \\
 & \Rightarrow f\left( x \right)\left( 1+{{x}^{2}} \right)=2x \\
 & \Rightarrow f\left( x \right)=\dfrac{2x}{1+{{x}^{2}}} \\
\end{align}$
We are asked the value of $f'\left( \dfrac{1}{2} \right)$.
We know that $\dfrac{d}{dx}\left( \dfrac{a\left( x \right)}{b\left( x \right)} \right)=\dfrac{b\left( x \right)\times a'\left( x \right)-a\left( x \right)\times b'\left( x \right)}{{{\left( b\left( x \right) \right)}^{2}}}$

Let us differentiate the value of the function $f\left( x \right)$, we get
$\begin{align}
  & f'\left( x \right)=\dfrac{d}{dx}\left( \dfrac{2x}{1+{{x}^{2}}} \right) \\
 & \Rightarrow f'\left( x \right)=\dfrac{\left( 1+{{x}^{2}} \right)\times \dfrac{d}{dx}\left( 2x \right)-\left( 2x \right)\dfrac{d}{dx}\left( 1+{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}=\dfrac{2\left( 1+{{x}^{2}} \right)-\left( 2x \right)\left( 2x \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \\
 & \Rightarrow f'\left( x \right)=\dfrac{2+2{{x}^{2}}-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}=\dfrac{2-2{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}} \\
\end{align}$
By substitute $x=\dfrac{1}{2}$, we get
$\begin{align}
  & f'\left( \dfrac{1}{2} \right)=\dfrac{2-2\times {{\left( \dfrac{1}{2} \right)}^{2}}}{{{\left( 1+{{\left( \dfrac{1}{2} \right)}^{2}} \right)}^{2}}}=\dfrac{2-2\times \dfrac{1}{4}}{{{\left( 1+\dfrac{1}{4} \right)}^{2}}}=\dfrac{\dfrac{3}{2}}{\dfrac{25}{16}}=\dfrac{3}{2}\times \dfrac{16}{25}=\dfrac{24}{25} \\
 & \Rightarrow f'\left( \dfrac{1}{2} \right)=\dfrac{24}{25} \\
\end{align}$
$\therefore $We get the required value as $f'\left( \dfrac{1}{2} \right)=\dfrac{24}{25}$.

So, the correct answer is “Option B”.

Note: Students must be careful while substituting the limits while using the Leibnitz rule of calculus. Some students forget to write the differentiation of the limit which is important when there are higher order terms as limits. In our question, we just have x as the limit in integration. Some students try to solve the integral equation given instead of differentiating it. That approach will not give the result. The key trick is that whenever we see a function of x or y as the limits in integration, we should use the Leibnitz rule.