#### Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 8

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

Hint:

Use quotient rule

Given:

$x=\frac{3 a t}{1+t^{2}} \\$      ,    $y=\frac{3 a t^{2}}{1+t^{2}} \\$

Solution:

$x=\frac{3 a t}{1+t^{2}} \\$

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

\begin{aligned} & &=\frac{\left(1+t^{2}\right) \frac{d(3 a t)}{d t}-3 a t \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}} \end{aligned}                                                        [Using quotient rule]

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

\begin{aligned} & &=\frac{3 a\left(1+t^{2}\right)-6 a t^{2}}{\left(1+t^{2}\right)^{2}} \end{aligned}

$=\frac{3 a+3 a t^{2}-6 a t^{2}}{\left(1+t^{2}\right)^{2}} \\$

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

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

$y=\frac{3 a t^{2}}{1+t^{2}} \\$

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

\begin{aligned} &\\ &=\frac{(1+t) \frac{d\left(3 a t^{2}\right)}{d t}-3 a t^{2} \frac{d\left(1+t^{2}\right)}{d t}}{\left(1+t^{2}\right)^{2}} \end{aligned}                                                                                                                     [Using quotient rule]

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

$=\frac{(1+t) \cdot 3 a \cdot(2 t)-3 a t^{2}(0+2 t)}{\left(1+t^{2}\right)}$                                                                     $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

$=\frac{6 a t\left(1+t^{2}\right)-6 a t^{3}}{\left(1+t^{2}\right)^{2}} \\$

$=\frac{6 a t+6 a t^{3}-6 a t^{3}}{\left(1+t^{2}\right)^{2}} \\$

\begin{aligned} & &\frac{d y}{d x}=\frac{6 a t}{(1+t)^{2}} \end{aligned}                                                                                                              (2)

Now

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

Put the value of   $\frac{dy}{dt}\:and\: \frac{dx}{dt}$  from the equation (2) and (1)

In  $\frac{dy}{dx}$

$\frac{d y}{d x}=\frac{\frac{6 a t}{\left(1+t^{2}\right)^{2}}}{\frac{3 a\left(1-t^{2}\right)}{\left(1+t^{2}\right)^{2}}}$

$=\frac{6 a t}{3 a\left(1-t^{2}\right)}$

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