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  • Please Solve RD Sharma Class 12 Chapter 10 Differentiation Exercise 10. 7 Question 12 Maths Textbook Solution.

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

Answers (1)

Answer:

            \frac{d y}{d x}=-1

Hint:

            Use chain rule

Given:

            \begin{aligned} &x=\cos ^{-1} \frac{1}{\sqrt{1+t^{2}}} \\ &y=\sin ^{-1} \frac{1}{\sqrt{1+t^{2}}} \\ &t \in R \end{aligned}

Solution:

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

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

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

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

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

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

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

Now

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

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

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

\frac{d y}{d x}=\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{1+t^{2}}}\right)^{2}}} \times \frac{-1}{2\left(1+t^{2}\right)^{\frac{3}{2}}} \times 2 t

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

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

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

\begin{aligned} &\ &\frac{d y}{d t}=\frac{-1}{\left(1+t^{2}\right)} \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 the equation (1) and (2) respectively

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

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

 

Posted by

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