Question

If $\sec \left(\frac{x+y}{x-y}\right)=a$ prove that $\frac{dy}{dx}=\frac{y}{x}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},\sec \left(\frac{x+y}{x-y}\right)=a \\ \Rightarrow \frac{x+y}{x-y}={\sec }^{-1}\left(a\right) \end{gathered}$$
Differentiate with respect to x, we get,
$$\begin{gathered} \Rightarrow \left[\frac{\left(x-y\right){\displaystyle \frac{d}{dx}}\left(x+y\right)-\left(x+y\right){\displaystyle \frac{d}{dx}}\left(x-y\right)}{{\left(x-y\right)}^2}\right]=0 \\ \Rightarrow \left(x-y\right)\left(1+\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)-\left(x+y\right)\left(1-\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)=0 \\ \Rightarrow \left(x-y\right)+\left(x-y\right)\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}-\left(x+y\right)+\left(x+y\right)\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=0 \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\left[x-y+x+y\right]=x+y-x+y \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\left(2x\right)=2y \\ \Rightarrow \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{y}{x} \end{gathered}$$