Topic of the Question:
The topic of this question is differential calculus, specifically focusing on logarithmic differentiation and the application of differentiation rules to implicit functions.
Step 1 : Understanding the Question:
We are given an equation where variables appear in both the base and the exponent: $y^x = e^{y - x}$. We are asked to find the derivative of $y$ with respect to $x$, denoted as $\frac{dy}{dx}$.
Step 2 : Key Formulas and Approach:
Logarithmic simplification: We take the natural logarithm ($\log$) of both sides of the equation to simplify the exponent using the identity $\log(a^b) = b \log a$.
Explicit expression: We rearrange the simplified equation to write $x$ explicitly as a function of $y$, i.e., $x = f(y)$.
Quotient rule: To find the derivative $\frac{dx}{dy}$, we use the formula: $\frac{d}{dy}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}$.
Reciprocal derivative identity: We obtain the final derivative using the relation $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$.
Step 3 : Detailed Explanation:
We start with the given implicit equation: $y^x = e^{y - x}$.
Taking the natural logarithm on both sides of this equation yields: $\log(y^x) = \log(e^{y - x})$.
Using the power rule of logarithms, we simplify this to: $x \log y = (y - x) \log e$.
Since the natural logarithm of $e$ is equal to 1 ($\log e = 1$), the equation becomes: $x \log y = y - x$.
To isolate $x$, we collect all terms containing $x$ on the left side of the equation: $x \log y + x = y$.
Factoring out $x$ gives: $x(1 + \log y) = y \implies x = \frac{y}{1 + \log y}$.
Next, we differentiate $x$ with respect to $y$ using the quotient rule, where $u = y$ and $v = 1 + \log y$:
This gives: $\frac{dx}{dy} = \frac{(1 + \log y) \cdot \frac{d}{dy}(y) - y \cdot \frac{d}{dy}(1 + \log y)}{(1 + \log y)^2}$.
Evaluating the individual derivatives: $\frac{dx}{dy} = \frac{(1 + \log y) \cdot 1 - y \cdot \left(\frac{1}{y}\right)}{(1 + \log y)^2}$.
Simplifying the numerator: $\frac{dx}{dy} = \frac{1 + \log y - 1}{(1 + \log y)^2} = \frac{\log y}{(1 + \log y)^2}$.
To find $\frac{dy}{dx}$, we take the reciprocal of our expression for $\frac{dx}{dy}$: $\frac{dy}{dx} = \frac{(1 + \log y)^2}{\log y}$.
Step 4 : Final Answer:
The derivative $\frac{dy}{dx}$ is equal to $\frac{(1 + \log y)^2}{\log y}$, which corresponds to Option (D).