Question

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

Hint:

Be extra vigilant when applying the chain rule to implicit logs like \( \log y \). Never forget to append a trailing factor of \( \frac{dy}{dx} \) due to function composition dependencies.

Answer 1

Concept: When a variable base is raised to a variable exponent power, \( [f(x)]^{g(y)} \), the derivative cannot be computed via direct simple power rules. Instead, we must use logarithmic differentiation by taking the natural logarithm (\( \log \)) of both sides to convert exponents into standard operational products, and then differentiate implicitly.

Step 1: Take the natural logarithm on both sides.
The given equation is: \[ (\sin x)^y = y^{\cos x} \] Taking \( \log \) on both sides: \[ \log\left((\sin x)^y\right) = \log\left(y^{\cos x}\right) \] Using power logarithm rules \( \log(M^N) = N\log M \): \[ y \log(\sin x) = \cos x \log y \]

Step 2: Differentiate both sides with respect to \( x \) using the product rule.
Differentiating the left side: \[ \frac{d}{dx}[y \cdot \log(\sin x)] = \frac{dy}{dx} \cdot \log(\sin x) + y \cdot \frac{1}{\sin x} \cdot \cos x = \frac{dy}{dx}\log(\sin x) + y\cot x \] Differentiating the right side: \[ \frac{d}{dx}[\cos x \cdot \log y] = (-\sin x)\log y + \cos x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = -\sin x\log y + \frac{\cos x}{y}\frac{dy}{dx} \] Equating both differentiated results: \[ \frac{dy}{dx}\log(\sin x) + y\cot x = -\sin x\log y + \frac{\cos x}{y}\frac{dy}{dx} \]

Step 3: Group terms containing \( \frac{dy}{dx} \) together to isolate it.
Collect all terms involving \( \frac{dy}{dx} \) on the left-hand side and remaining terms on the right-hand side: \[ \frac{dy}{dx}\log(\sin x) - \frac{\cos x}{y}\frac{dy}{dx} = -\sin x\log y - y\cot x \] Factoring out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left[ \log(\sin x) - \frac{\cos x}{y} \right] = -(\sin x\log y + y\cot x) \]

Step 4: Solve for \( \frac{dy}{dx} \) cleanly by taking a common denominator.
Simplify the expression inside the brackets: \[ \frac{dy}{dx} \left[ \frac{y\log(\sin x) - \cos x}{y} \right] = -(\sin x\log y + y\cot x) \] Now multiply both sides by \( y \) and divide by the brackets factor: \[ \frac{dy}{dx} = \frac{-y(\sin x\log y + y\cot x)}{y\log(\sin x) - \cos x} \] Distributing the negative sign through the denominator to make it clean: \[ \frac{dy}{dx} = \frac{y^2\cot x + y\sin x\log y}{\cos x - y\log(\sin x)} \]