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#### Need solution for RD Sharma Maths Class 12 Chapter 10 Differentiation Excercise 10.7 Question 7

$\frac{dy}{dx}=\frac{x}{y}$

Hint:

Use  $\frac{d\left(e^{x}\right)}{d x}=e^{x}$

Given:

$x=\frac{e^{t}+e^{-t}}{2} \\$ , $y=\frac{e^{t}-e^{-t}}{2} \$

Solution:

$x=\frac{e^{t}+e^{-t}}{2} \\$

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

$=\frac{1}{2}\left(e^{t}-e^{-t}\right) \\$

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

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

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

Now

$x=\frac{e^{t}+e^{-t}}{2} \\$

$2 x=e^{t}+e^{-t} \\$

\begin{aligned} & &e^{t}+e^{-t}=2 x \end{aligned}

Also

$y=\frac{e^{t}-e^{-t}}{2} \$

\begin{aligned} &\ &e^{t}-e^{-t}=2 y \end{aligned}
Put these values of  $\left(e^{t}+e^{-t}\right) \text { and }\left(e^{t}-e^{-t}\right)$   in $\frac{dy}{dx}$  expression

$\frac{d y}{d x}==\frac{e^{t}+e^{-t}}{e^{t}-e^{-t}}=\frac{2 x}{2 y}=\frac{x}{y}$