Question

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

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

Answer 1

Answer: (A)

$\because$ $\sec \left(\frac{x-y}{x+y}\right)=a$
$\Rightarrow$ $\frac{x-y}{x+y}={\sec }^{-1}a$ On differentiating w.r.t. $x,$ we get $\frac{(x+y)\left(1-\frac{dy}{dx}\right)-(x-y)\left(1+\frac{dy}{dx}\right)}{{(x+y)}^2}=0$
$\Rightarrow$ $x+y-x+y-\{x+y+x-y\}\frac{dy}{dx}=0$
$\Rightarrow$ $2y=2x\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{y}{x}$