The curve defined by $x^{2}y+y^3=4$ passes through the point $(1,1)$. (i) Find $\frac{dy}{dx}$ at $(1,1)$ using implicit differentiation. (ii) State whether the tangent at this point is horizontal, vertical, or neither.

Step-by-step solution

1. Differentiate both sides with respect to $x$: $\frac{d}{dx}(x^{2}y)+\frac{d}{dx}(y^3)=0$.
2. Apply product rule to $x^{2}y$: $2xy+x^2\frac{dy}{dx}$.
3. Differentiate $y^3$ using chain rule: $3y^2\frac{dy}{dx}$.
4. Combine terms: $2xy+x^2\frac{dy}{dx}+3y^2\frac{dy}{dx}=0$.
5. Factor $\frac{dy}{dx}$: $\frac{dy}{dx}(x^2+3y^2)=-2xy$.
6. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=-\frac{2xy}{x^2+3y^2}$.
7. Substitute the point $(1,1)$: $\frac{dy}{dx}=-\frac{2\cdot 1\cdot 1}{1^2+3\cdot 1^2}=-\frac{2}{4}=-\frac{1}{2}$.
8. Since the slope is finite and non‑zero, the tangent is neither horizontal nor vertical.

Concept tested: Implicit differentiation and tangent nature

Published 9 September 2026 · How this solution was prepared and checked