Maths · Differentiation of trigonometric, inverse trigonometric, logarithmic, exponential, composite and implicit functions
then
\( \boldsymbol{I} \boldsymbol{f} \quad \boldsymbol{x}^{\boldsymbol{y}}=\boldsymbol{e}^{\boldsymbol{x}-\boldsymbol{y}} \quad \) then
- A. \( \frac{d y}{d x} \) doesn't exist at \( x=1 \)
- B. \( \frac{d y}{d x}=0 \quad \) when \( \quad x=1 \)
- C. \( \frac{d y}{d x}=\frac{1}{2} \quad \) when \( \quad x=e \)
- D. None of these
Step-by-step solution
Given \(x^y = e^{x-y}\), take natural log: \(y \ln x = x - y\). Rearranging: \(y(\ln x + 1) = x\), so \(y = \frac{x}{\ln x + 1}\). Differentiate using quotient rule: \(\frac{dy}{dx} = \frac{(\ln x + 1) \cdot 1 - x \cdot \frac{1}{x}}{(\ln x + 1)^2} = \frac{\ln x}{(\ln x + 1)^2}\). At \(x=1\), \(\ln 1 = 0\), so \(\frac{dy}{dx} = 0\). At \(x=e\), \(\frac{dy}{dx} = \frac{1}{(1+1)^2} = \frac{1}{4}\), not \(\frac{1}{2}\). Thus option B is correct.