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  • Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 28

Explain solution RD Sharma class 12 Chapter 10 Differentiation Exercise 10.7 question 28

Answers (1)

Answer:

            \frac{d y}{d x}=1

Hint:

            Use product rule, quotient rule and   \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}

Given:

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

Solution:

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

Differentiate w.r.t t

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

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

=\frac{1}{\sqrt{1-\left(\frac{2 t}{1+t^{2}}\right)^{2}}} \times \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}}                                               [Using quotient rule]

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

=\frac{2+2 t^{2}-4 t^{2}}{\sqrt{1+t^{4}+2 t^{2}-4 t^{2}} \cdot\left(1+t^{2}\right)} \\

=\frac{2-2 t^{2}}{\sqrt{1+t^{4}-2 t^{2}} \cdot\left(1+t^{2}\right)} \\

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

\frac{d x}{d t}=\frac{2}{1+t^{2}} \\                                              (1)

\tan y=\frac{2 t}{1-t^{2}} \\                                                                                                                     

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

Differentiate w.r.t   t

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

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

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

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

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

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

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

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

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

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

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

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

\begin{aligned} & &\frac{d y}{d x}=1 \end{aligned}

 

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