Question

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

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

Answer 1

Answer: (B)

$\frac{y}{x}$

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

diff. both sides w.r.t. $x$

$\sec \left(\frac{x+y}{x-y}\right)\tan \left(\frac{x+y}{x-y}\right)\left[\frac{(x-y)[1+\frac{dy}{dx}]-(x+y)(1-\frac{dy}{dx})}{{(x-y)}^2}\right]=0$

$a\tan \left(\frac{x+y}{x-y}\right)\left[\frac{x-y+x\frac{dy}{dx}-y\frac{dy}{dx}-x-y+x\frac{dy}{dx}+y\frac{dy}{dx}}{{(x-y)}^2}\right]=0$

$a\tan \left(\frac{x+y}{x-y}\right)\left(\frac{2x\frac{dy}{dx}-2y}{{(x-y)}^2}\right)=0$

$a\tan \left(\frac{x+y}{x-y}\right)(x\frac{dy}{dx}-y)=0$

$\Rightarrow \frac{x+y}{x-y}=0$ or $x\frac{dy}{dx}-y=0$

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