Question

(a) If \( x^y = e^{x-y} \), then prove that \[ \frac{dy}{dx} = \frac{\log_e x}{(1 + \log_e x)^2}. \]

Hint:

Use logarithmic differentiation for equations involving powers and exponential terms.

Answer 1

Takingthelogarithmofbothsides: \[ y\log_ex=x-y. \] Differentiatingwithrespectto\(x\): \[ \log_ex\frac{dy}{dx}+\frac{y}{x}=1-\frac{dy}{dx}. \] Rearrangingterms: \[ \frac{dy}{dx}(1+\log_ex)=1-\frac{y}{x}. \] Fromtheoriginalequation\(y=\frac{x}{1+\log_ex}\),substitute\(y\): \[ \frac{dy}{dx}=\frac{\log_ex}{(1+\log_ex)^2}. \]