If \(\rm \sec \left(\frac{x-y}{x+y}\right)={a} \), then \(\rm \frac{d y}{d x} \) is

Given:

sec(\(\frac{x-y}{x+y}\)) = a

Concept Used:

Implicit differentiation.

Chain rule of differentiation.

Calculation:

⇒ \(\frac{1}{cos(\frac{x-y}{x+y})}\) = a

⇒ cos(\(\frac{x-y}{x+y}\)) = \(\frac{1}{a}\)

Differentiate both sides with respect to x:

⇒ -sin(\(\frac{x-y}{x+y}\)) × \(\frac{d}{dx}\)(\(\frac{x-y}{x+y}\)) = 0

⇒ -sin(\(\frac{x-y}{x+y}\)) × \(\frac{(x+y)(1-\frac{dy}{dx}) - (x-y)(1+\frac{dy}{dx})}{(x+y)^2}\) = 0

Since sin(\(\frac{x-y}{x+y}\)) is unlikely to be zero (as that would make 'a' undefined), we can focus on the other term:

⇒ (x+y)(1-\(\frac{dy}{dx}\)) - (x-y)(1+\(\frac{dy}{dx}\)) = 0

⇒ x + y - x\(\frac{dy}{dx}\) - y\(\frac{dy}{dx}\) - x + y - x\(\frac{dy}{dx}\) + y\(\frac{dy}{dx}\) = 0

⇒ 2y - 2x\(\frac{dy}{dx}\) = 0

⇒ 2x\(\frac{dy}{dx}\) = 2y

⇒ \(\frac{dy}{dx}\) = \(\frac{y}{x}\)

∴ \(\frac{dy}{dx}\) = \(\frac{y}{x}\)

Hence option 4 is correct