Question

If the value of \[y\sqrt{{{x}^{2}}+1}=\log \left( \sqrt{{{x}^{2}}+1}-x \right)\] then find the value of \[\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}+xy+1\]
\[\begin{align}
  & \text{A}.\text{ }0 \\
 & \text{B}.\text{ 1} \\
 & \text{C}.\text{ 2} \\
 & \text{D}.\text{ None of the above}. \\
\end{align}\]

Answer 1

Hint: To solve this question, we will differentiate the term on LHS and RHS of the given equation separately. While doing so, we will use $\dfrac{d}{dx}\left( x \right)=1,\dfrac{d}{dx}\left( \sqrt{x} \right)=\dfrac{1}{2\sqrt{x}}\text{ and }\dfrac{d}{dx}\left( \log \left( x \right) \right)=\dfrac{1}{x}$. Then, we will equate both differentiation above. Also, at the end we will use the chain rule of differentiation stated as \[\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\cdot g'\left( x \right)\]

Complete step-by-step answer:
We are given;
\[y\sqrt{{{x}^{2}}+1}=\log \left( \sqrt{{{x}^{2}}+1}-x \right)\]
We will separately calculate $\dfrac{dy}{dx}\text{ and }xy$
To calculate $\dfrac{dy}{dx}$ let us first calculate equation (i) with respect to x on both sides:
\[\dfrac{d}{dx}\left( y\sqrt{{{x}^{2}}+1} \right)=\dfrac{d}{dx}\left( \log \left( \sqrt{{{x}^{2}}+1}-x \right) \right)\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
To solve LHS of above equation we will use the product rule of differentiation stated as below:
\[\dfrac{d}{dx}\left( f\left( x \right)g\left( x \right) \right)=f\left( x \right)\dfrac{d}{dx}\left( g\left( x \right) \right)+g\left( x \right)\dfrac{d}{dx}\left( f\left( x \right) \right)\]
Where, f (x) and g (x) are function of x.
Consider LHS of equation (ii) as ${{I}_{1}}$ then
\[{{I}_{1}}=\dfrac{d}{dx}\left( y\sqrt{{{x}^{2}}+1} \right)\]
Differentiating using product rule of differentiation we get:
\[\Rightarrow {{I}_{1}}=y\dfrac{d}{dx}\left( \sqrt{{{x}^{2}}+1} \right)+\sqrt{{{x}^{2}}+1}\dfrac{d}{dx}\]
Now, using $\dfrac{d}{dt}\sqrt{t}=\dfrac{1}{2\sqrt{t}}$ in above we get:
\[\begin{align}
  & {{I}_{1}}=y\dfrac{1}{2\sqrt{{{x}^{2}}+1}}\left( 2x \right)+\sqrt{{{x}^{2}}+1}\dfrac{dy}{dx} \\
 & {{I}_{1}}=\dfrac{xy}{\sqrt{{{x}^{2}}+1}}+\sqrt{{{x}^{2}}+1}\dfrac{dy}{dx}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)} \\
\end{align}\]
Now, consider RHS of equation (ii) as ${{I}_{2}}$ we have:
\[{{I}_{2}}=\dfrac{d}{dx}\left( \log \left( \sqrt{{{x}^{2}}+1}-x \right) \right)\]
Using $\dfrac{d}{dx}\log x=\dfrac{1}{x}\text{ and }\dfrac{d}{dx}\left( x \right)=1$ in above also applying chain rule of differentiation stated as \[\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\cdot g'\left( x \right)\]
\[\begin{align}
  & {{I}_{2}}=\dfrac{1}{\left( \sqrt{{{x}^{2}}+1}-x \right)}\left( \dfrac{1}{2\sqrt{{{x}^{2}}+1}}\left( +2x \right)-\dfrac{d}{dx}\left( x \right) \right) \\
 & {{I}_{2}}=\dfrac{1}{\left( \sqrt{{{x}^{2}}+1}-x \right)}\left( \dfrac{x}{\sqrt{{{x}^{2}}+1}}-1 \right) \\
 & {{I}_{2}}=\dfrac{1}{\left( \sqrt{{{x}^{2}}+1}-x \right)}\left( \dfrac{-\left( -x+\sqrt{{{x}^{2}}+1} \right)}{\sqrt{{{x}^{2}}+1}} \right) \\
\end{align}\]
Cancelling the term $\left( \sqrt{{{x}^{2}}+1}-x \right)$ we get:
\[{{I}_{2}}=\dfrac{-1}{\sqrt{{{x}^{2}}+1}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iv)}\]
Now, by equation (ii) we had ${{I}_{1}}={{I}_{2}}$ so, equating ${{I}_{1}}\text{ and }{{I}_{2}}$ as obtained in equation (iii) and (iv) we get:
\[\begin{align}
  & {{I}_{1}}={{I}_{2}} \\
 & \Rightarrow \dfrac{xy}{\sqrt{{{x}^{2}}+1}}+\sqrt{{{x}^{2}}+1}\dfrac{dy}{dx}=\dfrac{-1}{\sqrt{{{x}^{2}}+1}} \\
\end{align}\]
Taking LCM of term on LHS of above equation:
\[\Rightarrow \dfrac{xy+\left( \sqrt{{{x}^{2}}+1} \right)\left( \sqrt{{{x}^{2}}+1} \right)\dfrac{dy}{dx}}{\sqrt{{{x}^{2}}+1}}=\dfrac{-1}{\sqrt{{{x}^{2}}+1}}\]
Cancelling $\sqrt{{{x}^{2}}+1}$ from both sides:
\[\begin{align}
  & xy+\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}=-1 \\
 & xy+\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}+1=0 \\
 & \left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}+xy+1=0 \\
\end{align}\]
Therefore, the value of \[\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}+xy+1=0\] so option A is correct.

Note: Students can get confused on the point where ${{I}_{2}}$ is calculated. It is as below:
\[{{I}_{2}}=\dfrac{d}{dx}\left( \log \left( \sqrt{{{x}^{2}}+1}-x \right) \right)\]
And as $\dfrac{d}{dt}\log t=\dfrac{1}{t}$
\[\begin{align}
  & {{I}_{2}}=\dfrac{1}{\sqrt{{{x}^{2}}+1}-x}\dfrac{d}{dx}\left( \sqrt{{{x}^{2}}+1}-x \right) \\
 & {{I}_{2}}=\dfrac{1}{\sqrt{{{x}^{2}}+1}-x}\left( \dfrac{1}{2\sqrt{{{x}^{2}}+1}}\dfrac{d}{dx}\left( {{x}^{2}}+1 \right)-\dfrac{d}{dx}x \right) \\
 & {{I}_{2}}=\dfrac{1}{\sqrt{{{x}^{2}}+1}-x}\left( \dfrac{2}{2\sqrt{{{x}^{2}}+1}}-1 \right) \\
 & {{I}_{2}}=\dfrac{1}{\sqrt{{{x}^{2}}+1}-x}\left( \dfrac{x}{\sqrt{{{x}^{2}}+1}}-1 \right) \\
\end{align}\]
This is how it is calculated.
Always remember that we apply chain rule of differentiation in such cases which is as below:
\[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right.\left( g'\left( x \right) \right)\]
In this question, we have automatically arrived at \[\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}+xy+1=0\] if this is not the case in any other question, then we will first calculate $\dfrac{dy}{dx}\text{ and }xy$ separately then add them by given condition of question to get result.