Resolved: "How do you find the slope of the..."

Dakota Livingston

Answered question

2022-04-10

How do you find the slope of the tangent to the curve

y^{3} x + y^{2} x^{2} = 6

at (2,1)?

Wrastirtyzp9w

Beginner2022-04-11Added 10 answers

Use implicit differentiation and the product rule

3 y^{2} \frac{d y}{d x} x + y^{3} + 2 y \frac{d y}{d x} x^{2} + y^{2} 2 x = 0

Do some rewriting

3 x y^{2} \frac{d y}{d x} + 2 x^{2} y \frac{d y}{d x} + y^{3} + 2 x y^{2} = 0

Factor and move terms without a

\frac{d y}{d x}

factor to right side

\frac{d y}{d x} (3 x y^{2} + 2 x^{2} y) = - y^{3} - 2 x y^{2}

now divide both sides by

3 x y^{2} + 2 x^{2} y

and factor where you can

\frac{d y}{d x} = \frac{- y^{2} (y + 2 x)}{y x (3 y + 2 x)}

\frac{d y}{d x} = \frac{- y (y + 2 x)}{x (3 y + 3 x)}

Now evaluate at the given point (2,1)

\frac{d y}{d x} = \frac{- 1 (1 + 2 (2))}{2 (3 (1) + 2 (2))} = \frac{- 1 (5)}{2 (7)} = \frac{- 5}{14}

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