Question

Find \(\frac{dy}{dx}\) when \[ (\sin y)^x = (\cos x)^y. \]

Hint:

Use logarithmic differentiation for equations with variables in both base and exponent, then apply implicit differentiation.

Answer 1

Given: \[ (\sin y)^x = (\cos x)^y. \] Take natural logarithm on both sides: \[ x \log(\sin y) = y \log(\cos x). \] Differentiate both sides with respect to \(x\), treating \(y\) as a function of \(x\): \[ \frac{d}{dx} \left( x \log(\sin y) \right) = \frac{d}{dx} \left( y \log(\cos x) \right). \] Using product rule on both sides: \[ \log(\sin y) + x \cdot \frac{1}{\sin y} \cdot \cos y \cdot \frac{dy}{dx} = \frac{dy}{dx} \cdot \log(\cos x) + y \cdot \frac{1}{\cos x} \cdot (-\sin x). \] Rewrite: \[ \log(\sin y) + x \cot y \frac{dy}{dx} = \frac{dy}{dx} \log(\cos x) - y \tan x. \] Group \(\frac{dy}{dx}\) terms on one side: \[ x \cot y \frac{dy}{dx} - \frac{dy}{dx} \log(\cos x) = - y \tan x - \log(\sin y). \] Factor \(\frac{dy}{dx}\): \[ \frac{dy}{dx} \left( x \cot y - \log(\cos x) \right) = - y \tan x - \log(\sin y). \] Therefore, \[ \boxed{ \frac{dy}{dx} = \frac{ - y \tan x - \log(\sin y) }{ x \cot y - \log(\cos x) }. } \]