Question

(b) Find \( \frac{dy}{dx} \), \text{ if } \[ 5^x + 5^y = 5^{x + y}. \]

Hint:

When differentiating exponential equations involving multiple variables in exponents, always apply the chain rule carefully and isolate \( \frac{dy}{dx} \) step by step. Simplify the equation methodically to avoid errors.

Answer 1

We are given the equation: \[ 5^x + 5^y = 5^{x + y}. \] To find \( \frac{dy}{dx} \), we will differentiate both sides with respect to \( x \) using implicit differentiation. Step 1: Differentiate the left-hand side. Using the chain rule, the derivatives of \( 5^x \) and \( 5^y \) are: \[ \frac{d}{dx} \left( 5^x \right) = 5^x \ln 5 \quad \text{and} \quad \frac{d}{dx} \left( 5^y \right) = 5^y \ln 5 \cdot \frac{dy}{dx}. \] So, the derivative of the left-hand side becomes: \[ 5^x \ln 5 + 5^y \ln 5 \cdot \frac{dy}{dx}. \] Step 2: Differentiate the right-hand side. For \( 5^{x + y} \), use the chain rule: \[ \frac{d}{dx} \left( 5^{x + y} \right) = 5^{x + y} \ln 5 \cdot \left( 1 + \frac{dy}{dx} \right). \] Step 3: Set up the equation. Equating the derivatives of the left-hand and right-hand sides: \[ 5^x \ln 5 + 5^y \ln 5 \cdot \frac{dy}{dx} = 5^{x + y} \ln 5 \cdot \left( 1 + \frac{dy}{dx} \right). \] Step 4: Simplify and solve for \( \frac{dy}{dx} \). Divide through by \( \ln 5 \) (since \( \ln 5 \neq 0 \)): \[ 5^x + 5^y \cdot \frac{dy}{dx} = 5^{x + y} \cdot \left( 1 + \frac{dy}{dx} \right). \] Expand the right-hand side: \[ 5^x + 5^y \cdot \frac{dy}{dx} = 5^{x + y} + 5^{x + y} \cdot \frac{dy}{dx}. \] Rearrange to isolate terms involving \( \frac{dy}{dx} \): \[ 5^y \cdot \frac{dy}{dx} - 5^{x + y} \cdot \frac{dy}{dx} = 5^{x + y} - 5^x. \] Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \cdot \left( 5^y - 5^{x + y} \right) = 5^{x + y} - 5^x. \] Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{5^{x + y} - 5^x}{5^y - 5^{x + y}}. \] Final Answer: \[ \boxed{\frac{dy}{dx} = \frac{5^{x + y} - 5^x}{5^y - 5^{x + y}}}. \]