Find dy/dx at (1, 2) for the curve x^2y + xy^2 = 6.

Derivatives & Second-Order Derivatives · Question 40 of 96

Question 40

MediumOriginal question

Find $\frac{dy}{dx}$ at $(1, 2)$ for the curve $x^2y + xy^2 = 6$ .

- \frac{8}{5}

Correct answerYour answer

- \frac{5}{8}

Correct answerYour answer

- \frac{3}{5}

Correct answerYour answer

\frac{8}{5}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

Product rule on each term: $2xy + x^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0$ .
At $(1, 2)$ : $4 + \frac{dy}{dx} + 4 + 4\frac{dy}{dx} = 0$ , so $5\frac{dy}{dx} = -8$ and $\frac{dy}{dx} = -\frac{8}{5}$ .

Worth noting

$-\frac{5}{8}$ divides the wrong way: the terms with $\frac{dy}{dx}$ add to $5\frac{dy}{dx}$ , and the rest to 8.

Ten questions from Derivatives & Second-Order Derivatives, at your own pace.

Ten questions from Derivatives & Second-Order Derivatives, at your own pace.