Question

If the value of $y\sqrt{{{x}^{2}}+1}=\log \left( \sqrt{{{x}^{2}}+1}-x \right)$ then find the value of $\left( {{x}^{2}}+1 \right)\dfrac{dy}{dx}+xy+1$

& \text{A}.\text{ }0 \\\ & \text{B}.\text{ 1} \\\ & \text{C}.\text{ 2} \\\ & \text{D}.\text{ None of the above}. \\\ \end{aligned}$$

Answer 1

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:

& {{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{aligned}$$ Now, consider RHS of equation (ii) as ${{I}_{2}}$ we have: $${{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{aligned} & {{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{aligned}$$ Cancelling the term $\left( \sqrt{{{x}^{2}}+1}-x \right)$ we get: $${{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{aligned} & {{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{aligned}$$ Taking LCM of term on LHS of above equation: $$\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{aligned} & 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{aligned}$$ Therefore, the value of $$\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{aligned} & {{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{aligned}$$ This is how it is calculated. Always remember that we apply chain rule of differentiation in such cases which is as below: $$\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.