In this post, we will learn a step-by-step approach to find the slope \( \left(\frac{dy}{dx} \right) \) for the curve defined by: \( x^2 y + y^3 = 6x + 1 \)
Solution:
\( x^2 y + y^3 = 6x + 1 \)
Differentiate both sides with respect to \( x \):
\( \displaystyle \frac{d}{dx}(x^2 y) + \frac{d}{dx}(y^3) = \frac{d}{dx}(6x) + \frac{d}{dx}(1) \\ \displaystyle \Rightarrow 2xy + x^2 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 6 \small \text{, applying the product rule to } x^2 y \text{ and the chain rule to } y^3 \\ \displaystyle \Rightarrow x^2 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 6\, – 2xy \\ \displaystyle \Rightarrow \frac{dy}{dx} (x^2 + 3y^2) = 6\, – 2xy \\ \displaystyle \Rightarrow \frac{dy}{dx} = \frac{6\, – 2xy}{x^2 + 3y^2} \\ \)Therefore, \( \frac{dy}{dx} \) for the curve defined by \( x^2 y + y^3 = 6x + 1 \) is
\[ \fbox{$ \frac{dy}{dx} = \frac{6\, – 2xy}{x^2 + 3y^2} $} \\ \]
Please let me know in the comments if you find any error in this solution.