Question

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

Answer 1

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,

& \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{aligned}$$ Here we have to apply chain rule with $$\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{aligned} & \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{aligned}$$ Now, we can write $$\dfrac{dy}{dx}$$ as $$\begin{aligned} & \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{aligned}$$ As we need to find the second derivative of the given function $$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{aligned} & \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{aligned}$$ Now, let us calculate $$\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{aligned} & 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{aligned}$$ Hence we can simply the above expression $$\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{aligned} & \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{aligned}$$ **Note:** Calculation is an important task for these kinds of questions otherwise one simple mistake will lead to a fully wrong answer and further calculation. We can apply $$\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.