#### Please Solve RD Sharma Class 12 Chapter 10 Differentiation Exercise 10.7 Question 21 Maths Textbook Solution.

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

Hint:

Use quotient rule and   $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Given:

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

Solution:

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

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

\begin{aligned} &\ &=a \times \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]

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

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

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

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

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

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

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

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

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

\begin{aligned} &\\ &\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} \end{aligned}

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

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

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

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