Question

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

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

Answer 1

Answer: (B)

$sec\left(\frac{x+y}{x-y}\right)=a$ $\Rightarrow$ $\left(\frac{x+y}{x-y}\right)={\sec }^{-1}(a)$ On differentiating with respect to $x$ $\Rightarrow$ $\frac{(x-y)\left(1+\frac{dy}{dx}\right)-(x+y)\left(1-\frac{dy}{dx}\right)}{{(x-y)}^2}=0$ $\Rightarrow$ $x+x\frac{dy}{dx}-y-y\frac{dy}{dx}-x+x\frac{dy}{dx}-y$ $+y\frac{dy}{dx}=0$ $\Rightarrow$ $2x\frac{dy}{dx}=2y$ $\Rightarrow$ $\frac{dy}{dx}=\frac{y}{x}$