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

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Find tangent lines using implicit differentiation

Full solution

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

x = -3y^3 + 3y^2 - 3

at

(-3,1)

.

Find Derivative with Respect to y: First, we need to find the derivative of $x$ with respect to $y$ , since the equation is given in terms of $y$ . We use the power rule for differentiation. $\newline$ $\frac{dx}{dy} = \frac{d}{dy}(-3y^3 + 3y^2 - 3)$ $\newline$ = $-9y^2 + 6y$
Evaluate Derivative at $y = 1$ : Next, we evaluate the derivative at the point $y = 1$ to find the slope of the tangent line at that point. $\newline$ $\frac{dx}{dy}$ at $y = 1$ = $-9(1)^2 + 6(1)$ $\newline$ = $-9 + 6$ $\newline$ = $-3$

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