Question

If \( x^{1/3} + y^{1/3} = 1 \), find \( \frac{dy}{dx} \) at the point \( \left( \frac{1}{8}, \frac{1}{8} \right) \).

Hint:

For implicit differentiation, always isolate \( \frac{dy}{dx} \) after applying the chain rule to each term.

Answer 1

Differentiate the given equation implicitly with respect to \( x \): \[ \frac{d}{dx} \left( x^{1/3} + y^{1/3} \right) = \frac{d}{dx}(1). \] Using the chain rule: \[ \frac{1}{3}x^{-2/3} + \frac{1}{3}y^{-2/3} \frac{dy}{dx} = 0. \] Rearrange to solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{x^{-2/3}}{y^{-2/3}}. \] Substitute \( x = \frac{1}{8} \) and \( y = \frac{1}{8} \): \[ x^{-2/3} = \left( \frac{1}{8} \right)^{-2/3} = \left( 8 \right)^{2/3} = 4, \quad y^{-2/3} = \left( \frac{1}{8} \right)^{-2/3} = \left( 8 \right)^{2/3} = 4. \] \[ \frac{dy}{dx} = -\frac{4}{4} = -1. \] Final Answer: \[ \boxed{-1} \]