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Find the slope of the curve: x^3 - 3xy^2 + y^3 = 1 at the point (2, -1).

Question:

Find the slope of the curve: {eq}x^3 - 3xy^2 + y^3 = 1 {/eq} at the point {eq}(2, \; -1) {/eq}.

The slope of a Tangent Line to a Curve:

If the tangent of the curve is: {eq}\ y=mx+b, {/eq}, then{eq}\ m= \dfrac{ \ dx}{ \ dy} {/eq} is the slope of the tangent line to the curve.

In this problem, we need to find out the slope of the tangent line to the curve at that specific point.

Answer and Explanation:

First, we need to find the slope of the tangent line to the curve {eq}x^3 - 3xy^2 + y^3 = 1 {/eq}

Differentiate both sides:

{eq}\Rightarrow \dfrac{...

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from PSAT Prep: Tutoring Solution

Chapter 10 / Lesson 13
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