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Find the slope of the tangent line to the graph of $5y^3 + 4 = x$ at $(-1,-1)$ .

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Find tangent lines using implicit differentiation

Full solution

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

5y^3 + 4 = x

at

(-1,-1)

.

Differentiate Implicitly: First, we need to find the derivative of the equation with respect to $x$ to get the slope of the tangent line. Since the equation is implicitly defined, we use implicit differentiation. Differentiate both sides with respect to $x$ : $\frac{d}{dx}(5y^3 + 4) = \frac{d}{dx}(x)$ $15y^2 \cdot \frac{dy}{dx} = 1$
Solve for $\frac{dy}{dx}$ : Next, solve for $\frac{dy}{dx}$ to find the slope of the tangent line at any point $(x, y)$ : $\frac{dy}{dx} = \frac{1}{15y^2}$
Substitute $y = -1$ : Now, substitute $y = -1$ into the derivative to find the specific slope at the point $(-1,-1$ : $\frac{dy}{dx} = \frac{1}{15(-1)^2}$ $\frac{dy}{dx} = \frac{1}{15}$

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