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} $