Question

If $x^{3}+2xy+\frac{1}{3}y^{3}=\frac{11}{3}$ then $\frac{dy}{dx}$ at $(2,-1)$ is

• A. -2
• B. 2
• C. 5
• D. -5
• E. -10

Hint:

Calculus Tip: When evaluating an implicit derivative at a specific point, plug the $(x,y)$ values in immediately after differentiating! Solving algebraically for $dy/dx$ first wastes time and increases the chance of algebra errors.

Answer 1

The Correct Option is A

Concept:
To find the derivative of an implicit function (where $y$ is not isolated on one side), differentiate both sides with respect to $x$ term-by-term. Use the Product Rule for mixed terms like $xy$, and use the Chain Rule for $y$ terms (attaching $\frac{dy}{dx}$ or $y^{\prime}$).

Step 1: Differentiate the equation term-by-term.
Given: $x^3 + 2xy + \frac{1}{3}y^3 = \frac{11}{3}$ Differentiate with respect to $x$: $$\frac{d}{dx}(x^3) + \frac{d}{dx}(2xy) + \frac{d}{dx}\left(\frac{1}{3}y^3\right) = \frac{d}{dx}\left(\frac{11}{3}\right)$$

Step 2: Apply the appropriate differentiation rules.
Apply the Product Rule on the middle term and Chain Rule on the third term: $$3x^2 + 2\left(x \frac{dy}{dx} + y(1)\right) + \frac{1}{3} \cdot 3y^2 \frac{dy}{dx} = 0$$

Step 3: Simplify the derivative equation.
Distribute the 2 and simplify the fractions: $$3x^2 + 2x \frac{dy}{dx} + 2y + y^2 \frac{dy}{dx} = 0$$ Let $y^{\prime} = \frac{dy}{dx}$ for easier notation: $$3x^2 + 2x y^{\prime} + 2y + y^2 y^{\prime} = 0$$

Step 4: Substitute the given coordinates immediately.
Rather than solving algebraically for $y^{\prime}$ first, substitute $x = 2$ and $y = -1$ immediately to simplify: $$3(2)^2 + 2(2)y^{\prime} + 2(-1) + (-1)^2 y^{\prime} = 0$$ $$3(4) + 4y^{\prime} - 2 + 1y^{\prime} = 0$$

Step 5: Solve for the numerical value of $y^{\prime$.}
Combine the constant terms and $y^{\prime}$ terms: $$12 - 2 + 5y^{\prime} = 0$$ $$10 + 5y^{\prime} = 0$$ $$5y^{\prime} = -10 \implies y^{\prime} = -2$$ Hence the correct answer is (A) -2.