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#### Please Solve RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 11 Maths Textbook Solution.

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

Hint:

Use quotient rule

Given:

\begin{aligned} x &=\frac{2 t}{1+t^{2}} \\ y &=\frac{1-t^{2}}{1+t^{2}} \end{aligned}

Solution:

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

\begin{aligned} & \\ &\frac{d x}{d t}=\frac{d\left(\frac{2 t}{1+t^{2}}\right)}{d t}=\frac{\left(1+t^{2}\right) \frac{d(2 t)}{d t}-2 t \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}} \end{aligned}                                   [Using quotient rule]

$\frac{d x}{d t}=\frac{\left(1+t^{2}\right)^{2} \times 2-2 t(2 t)}{\left(1+t^{2}\right)^{2}} \\$

$=\frac{2\left(1+t^{2}\right)-4 t^{2}}{\left(1+t^{2}\right)^{2}} \\$

$=\frac{2+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}} \\$

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

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

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

$=\frac{\left(1+t^{2}\right) \cdot \frac{d\left(1-t^{2}\right)}{d t}-\left(1-t^{2}\right) \cdot \frac{d\left(1+t^{2}\right)}{1+t}}{\left(1+t^{2}\right)}$                                             [Use quotient rule]

$=\frac{\left(1+t^{2}\right)(-2 t)-\left(1-t^{2}\right)(2 t)}{\left(1+t^{2}\right)^{2}} \\$

$\frac{d y}{d x}=\frac{-2 t-2 t^{3}-2 t+2 t^{3}}{\left(1+t^{2}\right)^{2}} \\$

\begin{aligned} &\frac{d y}{d t}=\frac{-4 t}{\left(1+t^{2}\right)^{2}} \end{aligned}                                                                                                             (2)

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

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

$\frac{d y}{d x}=\frac{\frac{-4 t}{\left(1+t^{2}\right)^{2}}}{\frac{\left(2-2 t^{2}\right)}{\left(1+t^{2}\right)^{2}}}$

$=\frac{-4 t}{2\left(1-t^{2}\right)}$

\begin{aligned} &\\ &=\frac{-2 t}{1-t^{2}} \end{aligned}

$\frac{d y}{d x}=-\frac{x}{y}$                                                                                             $\left[\frac{x}{y}=\frac{\frac{2 t}{1+t^{2}}}{\frac{1-t^{2}}{1+t^{2}}}=\frac{2 t}{1-t^{2}}\right]$