#### Provide Solution for RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 17

$\frac{d y}{d x}=\frac{x}{y}$

Hint:

Use  $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Given:

\begin{aligned} &x=a\left(t+\frac{1}{t}\right) \\ &y=a\left(t-\frac{1}{t}\right) \end{aligned}

Solution:

$x=a\left(t+\frac{1}{t}\right) \\$

\begin{aligned} & &\frac{d x}{d t}=a \frac{d\left(t+\frac{1}{t}\right)}{d t} \end{aligned}

\begin{aligned} &=a\left[\frac{d t}{d t}+\frac{d\left(\frac{1}{t}\right)}{d t}\right] \\ &=a\left[1+\frac{d(t)^{-1}}{d t}\right] \end{aligned}
$=a\left[1+\left\{(-1) t^{-1-1}\right\}\right] \\$                                                                                                                    $\left[\because \frac{d x^{n}}{d x}=n x^{n-1}\right]$

\begin{aligned} & &\frac{d x}{d t}=a\left(1-t^{2}\right) \end{aligned}                                                                                                                                                                  (1)

$y=a\left(t-\frac{1}{t}\right) \\$

$\frac{d y}{d t}=a \frac{d\left(t-\frac{1}{t}\right)}{d t} \\$

\begin{aligned} & &=a\left[\frac{d t}{d t}-\frac{d\left(\frac{1}{t}\right)}{d t}\right] \end{aligned}

$=a\left[1-\frac{d t^{-1}}{d t}\right] \\$

$=a\left[1-(-1) t^{-1-1}\right] \\$

\begin{aligned} & &\frac{d y}{d t}=a\left(1+t^{-2}\right) \end{aligned}                                                                                                                          (2)

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Put the values of  $\frac{d x}{d t} \text { and } \frac{d y}{d t}$  from the equation (1) and (2) respectively

$\frac{d y}{d x}=\frac{t^{2}+1}{t^{2}-1} \\$

\begin{aligned} & &=\frac{t\left(t+\frac{1}{t}\right)}{t\left(t-\frac{1}{t}\right)} \end{aligned}

$=\frac{\left(t+\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)} \\$

\begin{aligned} & &=\frac{a\left(t+\frac{1}{t}\right)}{a\left(t-\frac{1}{t}\right)} \end{aligned}                                                                                                        [Multiply and divide by a]

$=\frac{x}{y}$                                                                                                                       $\left[\because x=a\left(t+\frac{1}{t}\right) \& y=a\left(t-\frac{1}{t}\right)\right]$

Hence proved