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

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

Answers (1)

Answer:

            \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}

 

Posted by

infoexpert27

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