Question

If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

Hint:

Whenever variables appear in exponents, taking the natural logarithm on both sides immediately eliminates the exponential base \(e\), turning a complex implicit problem into a straightforward linear differential form.

Answer 1

Concept: This problem involves finding the derivative of an implicit function where the variable appears in both the base and the exponent. We can solve this either by using implicit differentiation directly or by taking the natural logarithm on both sides to simplify the exponential structure before differentiating.
• Logarithmic Property: \( \ln(e^k) = k \) and \( \ln(ab) = \ln a + \ln b \).
• Product Rule of Differentiation: \( \frac{d}{dx}[u \cdot v] = u \frac{dv}{dx} + v \frac{du}{dx} \).
• Chain Rule: Used for differentiating functions of \(y\) with respect to \(x\), i.e., \( \frac{d}{dx}[f(y)] = f'(y) \frac{dy}{dx} \).

Step 1: Simplifying the given equation by taking natural logarithms.
The given equation is: \[ xy = e^{x - y} \] Taking the natural logarithm (\(\ln\)) on both sides of the equation: \[ \ln(xy) = \ln\left(e^{x - y}\right) \] Using the logarithmic properties \( \ln(ab) = \ln a + \ln b \) and \( \ln(e^k) = k \), we can rewrite this as: \[ \ln x + \ln y = x - y \]

Step 2: Differentiating implicitly with respect to \(x\).
Now, we differentiate each term of the simplified equation with respect to \(x\): \[ \frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln y) = \frac{d}{dx}(x) - \frac{d}{dx}(y) \] Applying standard differentiation formulas (\(\frac{d}{dx}(\ln x) = \frac{1}{x}\) and applying the chain rule for \(\ln y\)): \[ \frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 1 - \frac{dy}{dx} \]

Step 3: Isolating the derivative term \( \frac{dy}{dx} \).
Collect all terms containing \(\frac{dy}{dx}\) on the left-hand side and the remaining terms on the right-hand side: \[ \frac{1}{y}\frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{1}{x} \] Factor out \(\frac{dy}{dx}\) from the left side: \[ \left(\frac{1}{y} + 1\right)\frac{dy}{dx} = \frac{x - 1}{x} \] Simplify the expression inside the parentheses: \[ \left(\frac{1 + y}{y}\right)\frac{dy}{dx} = \frac{x - 1}{x} \] Multiplying both sides by \(\frac{y}{1 + y}\) gives us the derivative in terms of \(x\) and \(y\): \[ \frac{dy}{dx} = \frac{y(x - 1)}{x(y + 1)} \]