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Find the slope of the tangent line to the graph of $y^3 - 1 = x$ at $(-2,-1)$ .

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Find tangent lines using implicit differentiation

Full solution

Q. Find the slope of the tangent line to the graph of

y^3 - 1 = x

at

(-2,-1)

.

Differentiate Implicitly: First, we need to differentiate the equation implicitly to find $\frac{dy}{dx}$ , which represents the slope of the tangent line. $\newline$ Differentiate both sides with respect to $x$ : $\newline$ $\frac{d}{dx}(y^3 - 1) = \frac{d}{dx}(x)$ $\newline$ $3y^2 \cdot \frac{dy}{dx} - 0 = 1$ $\newline$ $\frac{dy}{dx} = \frac{1}{(3y^2)}$
Substitute Given Point: Next, substitute the $y$ -coordinate of the given point $(-2, -1)$ into the derivative to find the slope at that point. $\newline$ Substitute $y = -1$ into $\frac{dy}{dx}$ : $\newline$ $\frac{dy}{dx} = \frac{1}{3(-1)^2}$ $\newline$ $\frac{dy}{dx} = \frac{1}{3}$

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