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Find the slope of the tangent line to the graph of $-y^3 + 4y^2 - 9 = x$ at $(-4,-1)$ .

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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 + 4y^2 - 9 = x

at

(-4,-1)

.

Find Derivative Using Implicit Differentiation: First, we need to find the derivative of the given equation with respect to $x$ to find the slope of the tangent line. Since the equation is implicitly defined, we use implicit differentiation. $\newline$ Differentiate both sides with respect to $x$ : $\newline$ $\frac{d}{dx}(-y^3 + 4y^2 - 9) = \frac{d}{dx}(x)$ $\newline$ Using the chain rule for the left side: $\newline$ $-3y^2\frac{dy}{dx} + 8y\frac{dy}{dx} = 1$
Solve for Slope: Next, solve for $\frac{dy}{dx}$ , which represents the slope of the tangent line. $\newline$ $-3y^2\left(\frac{dy}{dx}\right) + 8y\left(\frac{dy}{dx}\right) = 1$ $\newline$ Factor out $\frac{dy}{dx}$ : $\newline$ $\frac{dy}{dx}(-3y^2 + 8y) = 1$ $\newline$ $\frac{dy}{dx} = \frac{1}{(-3y^2 + 8y)}$
Substitute $y = -1$ : Now, substitute $y = -1$ into the derivative to find the specific slope at the point $(-4,-1)$ . $\newline$ $\frac{dy}{dx} = \frac{1}{-3(-1)^2 + 8(-1)}$ $\newline$ $= \frac{1}{-3 + -8}$ $\newline$ $= \frac{1}{-11}$

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