Question

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

Hint:

When differentiating terms with \(y\), don't forget to multiply by \(dy/dx\) (chain rule).
Using logarithms early on avoids dealing with complicated derivatives of product terms equal to exponential towers.

Answer 1

Concept:
• Implicit differentiation is used when \(y\) is not explicitly isolated.
• Taking natural logarithms on both sides can simplify expressions with variables in the exponent.

Step 1: Take natural logarithm on both sides
\[ \ln(xy) = \ln(e^{x-y}) \] Using log properties \(\ln(ab) = \ln a + \ln b\) and \(\ln(e^u) = u\):
\[ \ln x + \ln y = x - y \]

Step 2: Differentiate both sides with respect to \(x\)
\[ \frac{d}{dx}(\ln x) + \frac{d}{dx}(\ln y) = \frac{d}{dx}(x) - \frac{d}{dx}(y) \] \[ \frac{1}{x} + \frac{1}{y} \frac{dy}{dx} = 1 - \frac{dy}{dx} \]

Step 3: Isolate \(\frac{dy}{dx}\)
Move all terms with \(\frac{dy}{dx}\) to the left side:
\[ \frac{1}{y} \frac{dy}{dx} + \frac{dy}{dx} = 1 - \frac{1}{x} \] Factor out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \left(\frac{1}{y} + 1\right) = \frac{x - 1}{x} \] \[ \frac{dy}{dx} \left(\frac{1 + y}{y}\right) = \frac{x - 1}{x} \] Multiply both sides by \(\frac{y}{1 + y}\):
\[ \frac{dy}{dx} = \frac{y(x - 1)}{x(1 + y)} \]