To determine \(\frac{dy}{dx}\) from the equation \((\cos x)^y = (\sin y)^x\), we first apply the natural logarithm to both sides:
\(\ln((\cos x)^y) = \ln((\sin y)^x)\)
This simplifies to:
\(y \ln(\cos x) = x \ln(\sin y)\)
Next, we differentiate both sides with respect to \(x\). For the left side, using the product rule:
\(\frac{d}{dx}(y \ln(\cos x)) = y \cdot \frac{d}{dx}(\ln(\cos x)) + \ln(\cos x) \cdot \frac{dy}{dx}\)
The derivative of \(\ln(\cos x)\) is:
\(\frac{d}{dx}(\ln(\cos x)) = \frac{1}{\cos x} \cdot (-\sin x) = -\tan x\)
Therefore, the derivative of the left side is:
\(y (-\tan x) + \ln(\cos x) \cdot \frac{dy}{dx}\)
For the right side, applying the product rule:
\(\frac{d}{dx}(x \ln(\sin y)) = x \cdot \frac{d}{dx}(\ln(\sin y)) + \ln(\sin y) \cdot \frac{d}{dx}(x)\)
The derivative of \(\ln(\sin y)\) with respect to \(x\) is:
\(\frac{d}{dx}(\ln(\sin y)) = \frac{1}{\sin y}\cdot\cos y\cdot \frac{dy}{dx} = \cot y \cdot \frac{dy}{dx}\)
The derivative of the right side is thus:
\(x \cdot \cot y \cdot \frac{dy}{dx} + \ln(\sin y)\)
Equating the derivatives of both sides:
\(y (-\tan x) + \ln(\cos x) \cdot \frac{dy}{dx} = x \cdot \cot y \cdot \frac{dy}{dx} + \ln(\sin y)\)
Rearranging the terms to isolate \(\frac{dy}{dx}\):
\(\ln(\cos x) \cdot \frac{dy}{dx} - x \cdot \cot y \cdot \frac{dy}{dx} = \ln(\sin y) + y \tan x\)
Factoring out \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(\ln(\cos x) - x \cdot \cot y) = \ln(\sin y) + y \tan x\)
Solving for \(\frac{dy}{dx}\):
\(\frac{dy}{dx} = \frac{\ln(\sin y) + y \tan x}{\ln(\cos x) - x \cdot \cot y}\)
The final result is:
\(\frac{dy}{dx} = \frac{\log_e \sin y - y \tan x}{\log_e \cos x + x \cot y}\)