Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

Given the equation: $(\sin x)^y = y^{\cos x}$. Since the variables are in the base and the exponent, we apply the Logarithmic Differentiation method. Taking the natural logarithm ($\ln$) on both sides:

$$\ln((\sin x)^y) = \ln(y^{\cos x})$$

Using the logarithmic property $\ln(a^b) = b \ln a$, we simplify the expression to:

$$y \ln(\sin x) = \cos x \ln y$$

Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]

Differentiating both sides with respect to $x$ using the Product Rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$:

$$\frac{d}{dx}[y \ln(\sin x)] = \frac{d}{dx}[\cos x \ln y]$$

$$y \cdot \frac{1}{\sin x} \cdot \cos x + \ln(\sin x) \cdot \frac{dy}{dx} = \cos x \cdot \frac{1}{y} \cdot \frac{dy}{dx} + \ln y \cdot (-\sin x)$$

$$y \cot x + \ln(\sin x) \frac{dy}{dx} = \frac{\cos x}{y} \frac{dy}{dx} - \sin x \ln y$$

Grouping the $\frac{dy}{dx}$ terms on one side:

$$\frac{dy}{dx} \left[ \ln(\sin x) - \frac{\cos x}{y} \right] = - \sin x \ln y - y \cot x$$

Step 3 (Final Concluded Answer): [0.5 Mark]

Solving for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{-(\sin x \ln y + y \cot x)}{\ln(\sin x) - \frac{\cos x}{y}}$$

Multiplying numerator and denominator by $y$ to simplify:

$$\frac{dy}{dx} = \frac{y(y \cot x + \sin x \ln y)}{y \ln(\sin x) - \cos x}$$

💡 EL10 Concept Insight: When you see a variable "trapped" in an exponent, think of logarithms as a magic key that pulls the exponent down to the ground level, turning a difficult power problem into a much easier multiplication problem!