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

$\frac{d y}{d x}=-1$

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}