#### Please Solve RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 13 Maths Textbook Solution.

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

Hint:

Use quotient rule

Given:

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

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

Solution:

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

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

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

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

\begin{aligned} &\\ &=\frac{-2 t-2 t^{3}-2 t+2 t^{3}}{\left(1+t^{2}\right)^{2}} \end{aligned}

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

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

\begin{aligned} & &\frac{d y}{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]

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

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

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

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

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

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

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

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