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Find the slope of the tangent line to the graph of $4y^3 + 2 = x$ 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

4y^3 + 2 = x

at

(-2,-1)

.

Find Derivative: First, we need to find the derivative of the equation to get the slope formula. Since the equation is implicitly defined, we use implicit differentiation. Differentiate both sides with respect to $x$ : $\frac{d}{dx}(4y^3 + 2) = \frac{d}{dx}(x)$ $12y^2 \cdot \frac{dy}{dx} = 1$
Solve for $\frac{dy}{dx}$ : Next, solve for $\frac{dy}{dx}$ to find the slope of the tangent line: $\newline$ $\frac{dy}{dx} = \frac{1}{12y^2}$ $\newline$ Substitute $y = -1$ into the equation: $\newline$ $\frac{dy}{dx} = \frac{1}{12(-1)^2}$ $\newline$ $\frac{dy}{dx} = \frac{1}{12}$

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