Question

Given that \( x^y + y^x = a^b \), where \( a \) and \( b \) are positive constants, find \( \frac{dy}{dx} \).

Hint:

When differentiating terms with variables in both the base and the exponent, use the logarithmic rule and chain rule effectively.

Answer 1

Step 1: Differentiating the given equation. The given equation is: \[ x^y + y^x = a^b \] Let \( u = x^y \) and \( v = y^x \). Differentiating \( u \) and \( v \) with respect to \( x \), we have: \[ u = x^y \quad \Rightarrow \quad \frac{du}{dx} = x^{y-1}y + x^y \log x \cdot \frac{dy}{dx} \] \[ v = y^x \quad \Rightarrow \quad \frac{dv}{dx} = y^x \cdot \log y \cdot \frac{dy}{dx} + y^{x-1} \] Step 2: Differentiating the equation. Differentiating both sides of \( u + v = a^b \) with respect to \( x \), we get: \[ \frac{du}{dx} + \frac{dv}{dx} = 0 \] Substituting the values of \( \frac{du}{dx} \) and \( \frac{dv}{dx} \), we obtain: \[ x^{y-1}y + x^y \log x \cdot \frac{dy}{dx} + y^x \log y \cdot \frac{dy}{dx} + y^{x-1} = 0 \] Step 3: Simplifying the equation. Grouping terms involving \( \frac{dy}{dx} \), we get: \[ x^y \log x \cdot \frac{dy}{dx} + y^x \log y \cdot \frac{dy}{dx} = -x^{y-1}y - y^{x-1} \] Factoring out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left( x^y \log x + y^x \log y \right) = -\left( x^{y-1}y + y^{x-1} \right) \] Step 4: Solving for \( \frac{dy}{dx} \). \[ \frac{dy}{dx} = \frac{-\left( x^{y-1}y + y^{x-1} \right)}{x^y \log x + y^x \log y} \] Conclusion: The derivative \( \frac{dy}{dx} \) is: \[ \frac{dy}{dx} = \frac{-\left( x^{y-1}y + y^{x-1} \right)}{x^y \log x + y^x \log y}. \]