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} $