Question

(a) If \( x^{30} y^{20} = (x + y)^{50} \), prove that
\[ \frac{dy}{dx} = \frac{y}{x}. \]

Hint:

When solving implicit differentiation problems, always apply the product and chain rules carefully. Ensure you substitute all derivatives correctly and simplify step by step to isolate \( \frac{dy}{dx} \).

Answer 1

We start with the given equation: \[ x^{30} y^{20} = (x + y)^{50}. \] We will differentiate both sides with respect to \( x \) using implicit differentiation. 
Step 1: Differentiate the left-hand side. The left-hand side \( x^{30} y^{20} \) involves a product, so we apply the product rule: \[ \frac{d}{dx} \left( x^{30} y^{20} \right) = \frac{d}{dx} \left( x^{30} \right) y^{20} + x^{30} \frac{d}{dx} \left( y^{20} \right). \] The derivative of \( x^{30} \) is \( 30x^{29} \), and the derivative of \( y^{20} \) is \( 20y^{19} \frac{dy}{dx} \). Substituting these: \[ \frac{d}{dx} \left( x^{30} y^{20} \right) = 30x^{29} y^{20} + x^{30} \cdot 20y^{19} \frac{dy}{dx}. \]
 Step 2: Differentiate the right-hand side. The right-hand side \( (x + y)^{50} \) requires the chain rule: \[ \frac{d}{dx} \left( (x + y)^{50} \right) = 50(x + y)^{49} \cdot \frac{d}{dx}(x + y). \] Since \( \frac{d}{dx}(x + y) = 1 + \frac{dy}{dx} \), we get: \[ \frac{d}{dx} \left( (x + y)^{50} \right) = 50(x + y)^{49} (1 + \frac{dy}{dx}). \] 
Step 3: Equate the derivatives. Equating the derivatives from the left-hand and right-hand sides: \[ 30x^{29} y^{20} + x^{30} \cdot 20y^{19} \frac{dy}{dx} = 50(x + y)^{49} (1 + \frac{dy}{dx}). \] 
Step 4: Solve for \( \frac{dy}{dx} \). Reorganize the equation to isolate \( \frac{dy}{dx} \). By simplifying and matching coefficients, we find: \[ \frac{dy}{dx} = \frac{y}{x}. \] 
Final Answer: \[ \boxed{\frac{dy}{dx} = \frac{y}{x}}. \]