Question

If \[ (\sin x)^y=y^{\cos x}, \] then find \[ \frac{dy}{dx}. \] 

Hint:

Logarithmic differentiation is the only way to differentiate "variable to the power variable" functions. Be extremely careful with signs when moving terms from one side of the equation to the other.

Answer 1

Concept:
• For equations of the form \( f(x)^{g(y)} = h(y)^{k(x)} \), take the natural logarithm on both sides to bring the powers down. This is called logarithmic differentiation.
• Use the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \).
• Chain rule is essential for terms involving \(y\).

Step 1: Take the natural log of both sides
\[ \log(\sin x)^y = \log(y^{\cos x}) \] Using properties of logs: \[ y \log(\sin x) = \cos x \log y \]

Step 2: Differentiate both sides with respect to \(x\)
Differentiate LHS using product rule: \[ \frac{dy}{dx} \log(\sin x) + y \cdot \frac{d}{dx}(\log(\sin x)) = \frac{dy}{dx} \log(\sin x) + y \cdot \frac{1}{\sin x} \cdot \cos x \] \[ \text{LHS} = \frac{dy}{dx} \log(\sin x) + y \cot x \] Differentiate RHS using product rule: \[ -\sin x \log y + \cos x \cdot \frac{d}{dx}(\log y) = -\sin x \log y + \cos x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \] \[ \text{RHS} = -\sin x \log y + \frac{\cos x}{y} \frac{dy}{dx} \]

Step 3: Group terms containing \(dy/dx\)
\[ \frac{dy}{dx} \log(\sin x) - \frac{\cos x}{y} \frac{dy}{dx} = -\sin x \log y - y \cot x \] \[ \frac{dy}{dx} \left( \log(\sin x) - \frac{\cos x}{y} \right) = -(\sin x \log y + y \cot x) \]

Step 4: Isolate \(dy/dx\)
Multiply the left side bracket by \(y/y\): \[ \frac{dy}{dx} \left( \frac{y \log(\sin x) - \cos x}{y} \right) = -(\sin x \log y + y \cot x) \] \[ \frac{dy}{dx} = \frac{-y(\sin x \log y + y \cot x)}{y \log(\sin x) - \cos x} \]