What is the second derivative of $f\left( x \right)=\sin \left( {{x}^{2}} \right)$?

Hint: In this problem we need to calculate the second derivative of the given function that means we need to derive the given function two times with respect to the variable $x$. First, we will derive the given function with respect to the variable $x$ and simplify the equation by using the differentiation formulas $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$, $\dfrac{d}{dx}\left( \sin x \right)=\cos x$, $\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x$. By applying these formulas, we will get the value of ${{f}^{'}}\left( x \right)$ which is one time derivative of the given function. But we need to calculate the second derivative, so we will again derive the value ${{f}^{'}}\left( x \right)$ with respect to $x$. Here we will use the chain rule which is given by $\dfrac{d}{dx}\left( uv \right)=u{{v}^{'}}+v{{u}^{'}}$, some other differentiation formulas to simplify the equation.

Complete step-by-step answer:
Given function is $f\left( x \right)=\sin \left( {{x}^{2}} \right)$.
Differentiating the above function with respect to $x$, then we will get
${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( \sin \left( {{x}^{2}} \right) \right)$
Applying the differentiation formulas $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$, $\dfrac{d}{dx}\left( \sin x \right)=\cos x$, $\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x$ in the above equation, then we will get
$\begin{align}
  & {{f}^{'}}\left( x \right)=\cos \left( {{x}^{2}} \right)\times 2x \\
 & \Rightarrow {{f}^{'}}\left( x \right)=2x\cos \left( {{x}^{2}} \right) \\
\end{align}$
From the above equation we have the value of ${{f}^{'}}\left( x \right)$. Which is our one-time derivative of the given function. But we need to calculate the second derivative of the function. So, we are going to differentiate the value of ${{f}^{'}}\left( x \right)$ with respect to $x$, then we will have
${{f}^{''}}\left( x \right)=\dfrac{d}{dx}\left( 2x\cos \left( {{x}^{2}} \right) \right)$
Here $2$ which is in multiplication is a constant, so we can differentiate a constant in multiplication. So, take out the constant from the differentiation, then we will get
${{f}^{''}}\left( x \right)=2\dfrac{d}{dx}\left( x\cos \left( {{x}^{2}} \right) \right)$
Using the chain rule of the differentiation which is $\dfrac{d}{dx}\left( uv \right)=u{{v}^{'}}+v{{u}^{'}}$, then we will have
${{f}^{''}}\left( x \right)=2\left[ x\dfrac{d}{dx}\left( \cos \left( {{x}^{2}} \right) \right)+\cos \left( {{x}^{2}} \right)\times \dfrac{d}{dx}\left( x \right) \right]$
We know the value of $\dfrac{d}{dx}\left( x \right)=1$. Substituting this value in the above equation, then we will get
${{f}^{''}}\left( x \right)=2\left[ x\times \dfrac{d}{dx}\left( \cos \left( {{x}^{2}} \right) \right)+\cos \left( {{x}^{2}} \right) \right]$
Considering the value $\dfrac{d}{dx}\left( \cos \left( {{x}^{2}} \right) \right)$ separately and using the formula $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right)\times {{g}^{'}}\left( x \right)$ in the value, then we will get
$\dfrac{d}{dx}\left( \cos \left( {{x}^{2}} \right) \right)=\dfrac{d}{dx}\left( \cos \left( {{x}^{2}} \right) \right)\times \dfrac{d}{dx}\left( {{x}^{2}} \right)$
We know that $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, $\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x$. Substituting these values in the above equation, then we will have
$\begin{align}
  & \dfrac{d}{dx}\left( \cos \left( {{x}^{2}} \right) \right)=-\sin \left( {{x}^{2}} \right)\times 2x \\
 & \Rightarrow \dfrac{d}{dx}\left( \cos \left( {{x}^{2}} \right) \right)=-2x\sin \left( {{x}^{2}} \right) \\
\end{align}$
Now substituting the value of $\dfrac{d}{dx}\left( \cos \left( {{x}^{2}} \right) \right)$ in the value of ${{f}^{''}}\left( x \right)$, then we will get
$\begin{align}
  & {{f}^{''}}\left( x \right)=2\left[ x\left( -2x\sin \left( {{x}^{2}} \right) \right)+\cos \left( {{x}^{2}} \right) \right] \\
 & \Rightarrow {{f}^{''}}\left( x \right)=2\left[ -2{{x}^{2}}\sin \left( {{x}^{2}} \right)+\cos \left( {{x}^{2}} \right) \right] \\
 & \Rightarrow {{f}^{''}}\left( x \right)=-4{{x}^{2}}\sin \left( {{x}^{2}} \right)+2\cos \left( {{x}^{2}} \right) \\
\end{align}$

Hence the second derivative of the given function $f\left( x \right)=\sin \left( {{x}^{2}} \right)$ is ${{f}^{''}}\left( x \right)=-4{{x}^{2}}\sin \left( {{x}^{2}} \right)+2\cos \left( {{x}^{2}} \right)$.

Note: We can also solve this problem by using another method which is substitution method. We will use the substitution $u={{x}^{2}}$ and calculate the value of ${{f}^{'}}\left( x \right)$ in terms of $u$. Again, we will differentiate the value of ${{f}^{'}}\left( x \right)$ and calculate it in terms of $u$. In the final step again use the substitution $u={{x}^{2}}$ for the required result.