Question

If $\sec \left(\frac{x-y}{x+y}\right)=a$, then $\frac{dy}{dx}$is equal to

• A. $\frac{y}{x}$
• B. $\frac{-y}{x}$
• C. $\frac{x}{y}$
• D. $\frac{-x}{y}$

Answer 1

The correct option is A

$\frac{y}{x}$

Explanation for the correct option:

$\sec \left(\frac{x-y}{x+y}\right)=a....\left(1\right)$

Differentiate the equation $\left(1\right)$ with respect to $x$

$\begin{array}{rcl}& \Rightarrow & \sec \left(\frac{x-y}{x+y}\right)\tan \left(\frac{x-y}{x+y}\right)\left[\frac{\left(x+y\right)\left\{1-{\displaystyle \frac{dy}{dx}}\right\}-\left(x-y\right)\left\{1+{\displaystyle \frac{dy}{dx}}\right\}}{{\left(x+y\right)}^2}\right]=0\\ & \Rightarrow & a\tan \left(\frac{x-y}{x+y}\right)\left[\frac{x+y-x{\displaystyle \frac{dy}{dx}}-y{\displaystyle \frac{dy}{dx}}-x+y-x{\displaystyle \frac{dy}{dx}}+y{\displaystyle \frac{dy}{dx}}}{{\left(x+y\right)}^2}\right]=0\\ & \Rightarrow & a\tan \left(\frac{x-y}{x+y}\right)\left(\frac{-2x{\displaystyle \frac{dy}{dx}}+2y}{{\left(x+y\right)}^2}\right)=0\\ & \Rightarrow & a\tan \left(\frac{x-y}{x+y}\right)\left(-x\frac{dy}{dx}+y\right)=0\end{array}$

$\begin{array}{rcl}& \Rightarrow & \frac{x-y}{x+y}=0\mathrm{and}-x\frac{dy}{dx}+y=0\\ & \Rightarrow & -x\frac{dy}{dx}=-y\\ & \Rightarrow & \frac{dy}{dx}=\frac{y}{x}\end{array}$

Hence, option $\left(\mathit{A}\right)$ is correct.