Question

If $\sqrt{x^2+y^2}=a{e}^{{\tan }^{-1}\left(\frac{y}{x}\right)}$, $a>0,y\left(0\right)\ne 0$ then $f\text{'}\text{'}\left(0\right)$ is:

• A. $\left(\frac{a}{e}\right)e^{-\mathrm{\pi }/2}$
• B. $a{e}^{\mathrm{\pi }/2}$
• C. $\frac{-2e^{-\mathrm{\pi }/2}}{a}$
• D. $\left(\frac{a}{e}\right)e^{\mathrm{\pi }/2}$

Answer 1

The correct option is C

$\frac{-2e^{-\mathrm{\pi }/2}}{a}$

Explanation for the correct option:

Second order derivative:

In $\sqrt{x^2+y^2}=a{e}^{{\tan }^{-1}\left(\frac{y}{x}\right)}$, by taking log both sides, we get

$\begin{array}{rcl}\log \sqrt{x^2+y^2}& =& \log \left(a{e}^{{\tan }^{-1}\left(\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)}\right)\\ & \Rightarrow & \frac{1}{2}\log \left(x^2+y^2\right)=\log a+\log e^{{\tan }^{-1}\left(\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)}\\ & \Rightarrow & \frac{1}{2}\log \left(x^2+y^2\right)=\log a+{\tan }^{-1}\left(\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.\right)\end{array}$

By differentiating both sides w.r.t $x$, we get

$$\begin{gathered} \frac{1}{2}\times \frac{1}{x^2+y^2}\times \left(2x+2yy\text{'}\right)=\frac{1}{1+{\left({\displaystyle \frac{y}{x}}\right)}^2}\times \frac{xy\text{'}-y}{x^2} \\ \Rightarrow \frac{x+yy\text{'}}{x^2+y^2}=\frac{xy\text{'}-y}{x^2+y^2} \\ \Rightarrow x+yy\text{'}=xy\text{'}-y..........\left(1\right) \\ \Rightarrow y\text{'}=\frac{x+y}{x-y} \\ \Rightarrow y\text{'}\left(0\right)=-1 \end{gathered}$$

By differentiating $\left(1\right)$ w.r.t $x$, we get

$$\begin{gathered} 1+yy\text{'}\text{'}+y{\text{'}}^2=xy\text{'}\text{'}+y\text{'}-y\text{'} \\ \Rightarrow y\text{'}\text{'}=\frac{1+y{\text{'}}^2}{x-y} \\ \Rightarrow y\text{'}\text{'}\left(0\right)=\frac{1+1}{a{e}^{\mathrm{\pi }/2}}\left[\because y\left(0\right)=a{e}^{\pi /2}\right] \\ \Rightarrow y\text{'}\text{'}\left(0\right)=\frac{2e^{-\mathrm{\pi }/2}}{a} \end{gathered}$$

Hence, option C is correct.