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

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

Answers (1)

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Answer:

            \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}

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