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Find the slope of the tangent line to the graph of $6y^3 - 7 = x$ 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

6y^3 - 7 = x

at

(-1,1)

.

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. $\newline$ Differentiate both sides with respect to $x$ : $\newline$ $\frac{d}{dx}(6y^3 - 7) = \frac{d}{dx}(x)$ $\newline$ $18y^2 \cdot \frac{dy}{dx} = 1$ $\newline$ Solving for $\frac{dy}{dx}$ gives: $\newline$ $\frac{dy}{dx} = \frac{1}{(18y^2)}$
Implicit Differentiation: Next, substitute $y = 1$ into the derivative formula to find the slope at the point $(-1,1)$ . $\newline$ $\frac{dy}{dx}$ at $y = 1$ is: $\newline$ $\frac{dy}{dx} = \frac{1}{18 \times 1^2}$ $\newline$ $\frac{dy}{dx} = \frac{1}{18}$

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