The given function is $(\cos x)^y$ = $(\cos y)^x$ Taking logarithm on both the sides, we obtain ylog cosx = xlog cosy Differentiating both sides, we obtain log cosx × $\frac{dy}{dx}$ + y × $\frac{d}{dx}$ (log cos x) = log cos y × $\frac{d}{dx}$ (x) + x × $\frac{d}{dx}$ (log cos y) ⇒ log cos x × $\frac{dy}{dx}$ + y × $\frac{1}{\cos x}$ × $\frac{d}{dx}$ (cos x) = log cos y × 1 + x × $\frac{1}{\cos y}$ × $\frac{d}{dx}$ (cos y) ⇒ log cos x × $\frac{dy}{dx}$ + $\frac{y}{\cos x}$ (- sin x) = log cos y + $\frac{x}{\cos y}$ × (- sin y) × $\frac{dy}{dx}$ ⇒ log cos x × $\frac{dy}{dx}$ - y tan x - log cos y - x tan y × $\frac{dy}{dx}$ ⇒ log cos x × $\frac{dy}{dx}$ + x tan y × $\frac{dy}{dx}$ = log cos y + y tan x ⇒ (log cos x + x tan y) × $\frac{dy}{dx}$ = log cos y + y tan x ∴ $\frac{dy}{dx}$ = $\frac{\log\cos y + y\tan x}{\log\cos x + x\tan y}$ OR We have, siny = x sin (a + y) ⇒ x = $\frac{\sin y}{\sin(a+y)}$ Differentiating the above function we have, 1 = ⇒ $\sin^2$ (a + y) = [sin (a + y) × cos y - sin y cos (a + y)] $\frac{dy}{dx}$ ⇒ = $\frac{dy}{dx}$ ⇒ $\frac{\sin^2(a+y)}{\sin(a+y-y)}$ = $\frac{dy}{dx}$ ⇒ $\frac{\sin^2(a+y)}{\sin a}$ = $\frac{dy}{dx}$ ⇒ $\frac{dy}{dx}$ = $\frac{\sin^2(a+y)}{\sin a}$