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

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

Answers (1)

Answer:

            \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]

Posted by

infoexpert27

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