Given, $y = z^x$, then $y = z^y$. Find $\frac{dy}{dx}$.
Step-by-Step Solution
Key Concept: For equations of the form $y = z^x$ with additional constraints, take logarithms and differentiate implicitly to find relationships.
Taking $\ln$ on both sides of $\ln y = y\ln z$. Differentiating: $\frac{1}{y}\frac{dy}{dx} = y\frac{1}{dx}\ln z + \ln z \frac{dy}{dx}$. Rearranging: $\frac{d}{dx}[y\ln z] = \frac{z^y}{x(1-y\ln z)}$ after substituting the relationship between $y$, $z$, and $x$. The final derivative is $\frac{dy}{dx} = \frac{y^3}{x(1-y\ln z)}$.
Correct Answer: $\frac{y^3}{x(1-y\ln z)}$