Question

Find the second derivative of the following function.
\[y={{x}^{2}}\sqrt{1+{{x}^{2}}}\]

Answer 1

Hint: Apply multiplication of rule of differentiation and take care of chain rule as well whenever required. Both are given as,
Chain Rule: $\left( f\left( g\left( x \right) \right) \right)'=f'\left( g\left( x \right) \right).g'\left( x \right)$
Multiplication Rule of differentiation: -
$\dfrac{d}{dx}\left( u\left( x \right).v\left( x \right) \right)=u\left( x \right)\dfrac{dv\left( x \right)}{dx}+v\left( x \right)\dfrac{du\left( x \right)}{dx}$

Complete step by step answer:
We have the function
\[y={{x}^{2}}\sqrt{1+{{x}^{2}}}\] -(1)
Here, we can observe that \[y={{x}^{2}}\sqrt{1+{{x}^{2}}}\] has two functions \[{{x}^{2}}\] and \[{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}\] in multiplication. So, for the differentiation of function ‘y’ we need to apply multiplication rule of differentiation as stated below: -
If we have two functions u(x) and v(x) in multiplication as
\[y=u\left( x \right)v\left( x \right)\]
Then we can differentiate the above expression in following manner: -
\[\dfrac{dy}{dx}=\dfrac{d}{dx}\left( u\left( x \right)v\left( x \right) \right)=u\left( x \right)\dfrac{dv\left( x \right)}{dx}+v\left( x \right)\dfrac{du\left( x \right)}{dx}-(2)\]
Now, coming to the question or equation (1)
We can observe that we have \[u={{x}^{2}}\] and \[v=\sqrt{1+{{x}^{2}}}\] from equation (2)
Hence,
\[\begin{align}
  & \dfrac{d}{dx}\left( {{x}^{2}}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right)={{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}\dfrac{d}{dx}\left( {{x}^{2}} \right)+{{x}^{2}}\dfrac{d}{dx}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \\
 & \dfrac{dy}{dx}={{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}2x+{{x}^{2}}\dfrac{d}{dx}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \\
\end{align}\]
Here we have to apply chain rule with \[\dfrac{d}{dx}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}\] as stated below: -
If we have an implicit function like two or more function involved as \[f\left( g\left( x \right) \right)\] then differentiation of \[f\left( g\left( x \right) \right)\] is done in following manner: -
\[f{{\left( g\left( x \right) \right)}^{'}}={{f}^{'}}\left( g\left( x \right) \right){{g}^{'}}\left( x \right)-(3)\]
Hence,
\[\begin{align}
  & \dfrac{d}{dx}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}=\dfrac{1}{2}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}-1}}\dfrac{d}{dx}\left( {{x}^{2}} \right)\left( \because \dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}} \right) \\
 & \dfrac{d}{dx}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}=\dfrac{1}{2}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}}2x \\
 & \dfrac{d}{dx}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}=x{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}} \\
\end{align}\]
Now, we can write \[\dfrac{dy}{dx}\] as
\[\begin{align}
  & \dfrac{dy}{dx}=2x\sqrt{1+{{x}^{2}}}+\dfrac{{{x}^{2}}.x}{\sqrt{1+{{x}^{2}}}} \\
 & \dfrac{dy}{dx}=2x\sqrt{1+{{x}^{2}}}+\dfrac{{{x}^{3}}}{\sqrt{1+{{x}^{2}}}}-(4) \\
\end{align}\]
As we need to find the second derivative of the given function \[y={{x}^{2}}\sqrt{1+{{x}^{2}}}\] .
Hence, \[\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2x{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}+{{x}^{3}}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}} \right)\]
Therefore, we can write \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] as
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 2x{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}+{{x}^{3}}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}} \right)-(5)\]
Let us calculate \[\dfrac{d}{dx}\left( 2x{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right)\] and \[\dfrac{d}{dx}\left( {{x}^{3}}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}} \right)\] one by one.
\[\dfrac{d}{dx}\left( 2x{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right)=2\dfrac{d}{dx}\left( x{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right)\]
Here, we can observe that \[x\sqrt{1+x{}^{2}}\] is multiplication two functions x and \[{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}\] . So, here we need to apply multiplication rule of differentiation as stated in equation (2) where
\[u=x,v={{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}\]
Here,
\[\begin{align}
  & \dfrac{d}{dx}\left( x{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right)=x\dfrac{d}{dx}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}\dfrac{d}{dx}\left( x \right) \\
 & \dfrac{d}{dx}\left( x{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right)=x\times {{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}-1}}\dfrac{d}{dx}\left( {{x}^{2}} \right)+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}\left( \because \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right) \\
 & \dfrac{d}{dx}\left( x{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}} \right)=\dfrac{{{x}^{2}}}{\sqrt{1+{{x}^{2}}}}+\sqrt{1+{{x}^{2}}}-(6) \\
 & \\
\end{align}\]
Now, let us calculate \[\dfrac{d}{dx}\left( {{x}^{3}}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}} \right)\] where we can observe that two functions \[{{x}^{3}}\] and \[{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}}\] are in multiplication. So, here we also need to apply multiplication rule of differentiation as stated in equation (2) where
\[\begin{align}
  & u={{x}^{3}},v={{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}} \\
 & \dfrac{d}{dx}\left( {{x}^{3}}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}} \right)={{x}^{3}}\dfrac{d}{dx}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}}+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}}\dfrac{d}{dx}\left( {{x}^{3}} \right)={{x}^{3}}\times \left( \dfrac{-1}{2} \right){{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}-1}}\dfrac{d}{dx}\left( {{x}^{2}} \right)+{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}}\left( 3{{x}^{2}} \right)\left( \because \dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}} \right) \\
\end{align}\]
Hence we can simply the above expression
\[\dfrac{d}{dx}\left( {{x}^{3}}{{\left( 1+{{x}^{2}} \right)}^{\dfrac{-1}{2}}} \right)=\dfrac{-{{x}^{4}}}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{3}{2}}}}+\dfrac{3{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}}-(7)\]
Now, put values of equation (6) and (7) in equation (5) in following way: -
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2\left( \dfrac{{{x}^{2}}}{\sqrt{1+{{x}^{2}}}}+\sqrt{1+{{x}^{2}}} \right)-\dfrac{{{x}^{4}}}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{3}{2}}}}+\dfrac{3{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{\dfrac{1}{2}}}}\]
Hence,
\[\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2{{x}^{2}}\left( 1+{{x}^{2}} \right)+{{\left( 1+{{x}^{2}} \right)}^{2}}-{{x}^{4}}+3{{x}^{2}}\left( 1+{{x}^{2}} \right)}{\left( 1+{{x}^{2}} \right)\sqrt{1+{{x}^{2}}}} \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2{{x}^{2}}+2{{x}^{4}}+2+2{{x}^{4}}+4{{x}^{2}}-{{x}^{4}}+3{{x}^{2}}+3{{x}^{4}}}{\left( 1+{{x}^{2}} \right)\sqrt{1+{{x}^{2}}}} \\
 & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{6{{x}^{4}}+5{{x}^{2}}+2}{\left( 1+{{x}^{2}} \right)\sqrt{1+{{x}^{2}}}} \\
\end{align}\]

Note:
Calculation is an important task for these kinds of questions otherwise one simple mistake will lead to a fully wrong answer and further calculation.
We can apply \[\left( \dfrac{u}{v} \right)\] rule in place of uv rule and vice-versa is also true by just taking denominator to numerator or numerator to denominator as following way: -
Ex: - If we have \[\dfrac{d}{dx}\left( \sin x.x \right)\] we can write \[\dfrac{d}{dx}\left( \dfrac{\sin x}{\left( {{x}^{-1}} \right)} \right)\] and can apply \[\left( \dfrac{u}{v} \right)\] rule and let have to calculate \[\dfrac{d}{dx}\left( \dfrac{\sin x}{\left( 1+{{x}^{2}} \right)} \right)\] then we can write it as \[\sin x{{\left( 1+{{x}^{2}} \right)}^{-1}}\] and apply (u.v) rule. Hence, don’t get confused with (u.v) and \[\left( \dfrac{u}{v} \right)\] rules.
One can go wrong while applying chain rule as stated in equation (3). We need to apply it carefully to get further correct solutions.