Question

If \(y^x = e^{y - x}\), then \(\frac{dy}{dx}\) is equal to

• A. \( \frac{1 + \log y}{y \log y} \)
• B. \( \frac{(1 + \log y)^2}{y \log y} \)
• C. \( \frac{1 + \log y}{(\log y)^2} \)
• D. \( \frac{(1 + \log y)^2}{\log y} \)

Hint:

Whenever an implicit function can be explicitly rewritten as \(x = f(y)\), finding \(\frac{dx}{dy}\) first via the quotient rule is often much less prone to algebraic errors than performing implicit differentiation with the product rule directly on \(x \log y = y - x\).

Answer 1

The Correct Option is D

Concept: When a variable exists in the exponent of a function, we apply logarithmic differentiation. Taking the natural logarithm (\(\log_e\) or \(\ln\)) on both sides simplifies the exponents into linear products using the logarithm power rule: \[ \log(a^b) = b \log a \] Once simplified, we can express \(x\) explicitly in terms of \(y\) to compute \(\frac{dx}{dy}\), and then find the derivative using the reciprocal identity: \[ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} \] Step 1: Taking the natural logarithm on both sides and simplifying.
Given equation: \[ y^x = e^{y - x} \] Taking natural log (\(\log\)) on both sides: \[ \log(y^x) = \log(e^{y - x}) \] \[ x \log y = (y - x) \log e \] Since \(\log e = 1\), the expression simplifies to: \[ x \log y = y - x \] Rearranging terms to collect all \(x\) parameters on one side: \[ x \log y + x = y \] \[ x(1 + \log y) = y \quad \Rightarrow \quad x = \frac{y}{1 + \log y} \quad \cdots (1) \]

Step 2: Differentiating \(x\) with respect to \(y\) using the Quotient Rule.
Recall the quotient rule for differentiation: \(\frac{d}{dy}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}\). Applying this to equation (1): \[ \frac{dx}{dy} = \frac{(1 + \log y) \cdot \frac{d}{dy}(y) - y \cdot \frac{d}{dy}(1 + \log y)}{(1 + \log y)^2} \] \[ \frac{dx}{dy} = \frac{(1 + \log y) \cdot 1 - y \cdot \left(0 + \frac{1}{y}\right)}{(1 + \log y)^2} \] \[ \frac{dx}{dy} = \frac{1 + \log y - 1}{(1 + \log y)^2} = \frac{\log y}{(1 + \log y)^2} \]

Step 3: Finding \(\frac{dy}{dx}\).
Taking the reciprocal to get the final derivative: \[ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{(1 + \log y)^2}{\log y} \]