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If $\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k$ (a constant), then $\frac{d y}{d x}$ is equal to
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The correct answer is:
$\frac{y}{x}$
$\begin{aligned}
& \text { Given, } \cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k \\
& \Rightarrow \quad \frac{x^2-y^2}{x^2+y^2}=\cos k
\end{aligned}$
On differentiating w.r.t. $x$, we get
$\begin{array}{r}\frac{\left(x^2+y^2\right)\left(2 x-2 y \frac{d y}{d x}\right)-\left(x^2-y^2\right)\left(2 x+2 y \frac{d y}{d x}\right)}{\left(x^2+y^2\right)^2} \\ =0\end{array}$
$\begin{aligned} \Rightarrow & -4 x^2 y \frac{d y}{d x}+4 x y^2 =0 \\ \Rightarrow & \frac{d y}{d x} =\frac{y}{x}\end{aligned}$
& \text { Given, } \cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k \\
& \Rightarrow \quad \frac{x^2-y^2}{x^2+y^2}=\cos k
\end{aligned}$
On differentiating w.r.t. $x$, we get
$\begin{array}{r}\frac{\left(x^2+y^2\right)\left(2 x-2 y \frac{d y}{d x}\right)-\left(x^2-y^2\right)\left(2 x+2 y \frac{d y}{d x}\right)}{\left(x^2+y^2\right)^2} \\ =0\end{array}$
$\begin{aligned} \Rightarrow & -4 x^2 y \frac{d y}{d x}+4 x y^2 =0 \\ \Rightarrow & \frac{d y}{d x} =\frac{y}{x}\end{aligned}$
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