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Find the slope of the tangent line to the graph of $x = -4y^2 + 2y + 13$ at $(1,2)$ .

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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 = -4y^2 + 2y + 13

at

(1,2)

.

Find Derivative of x: First, we need to find the derivative of $x$ with respect to $y$ , since the equation is implicitly defined. We differentiate each term separately: $\newline$ $\frac{dx}{dy} = \frac{d}{dy}(-4y^2) + \frac{d}{dy}(2y) + \frac{d}{dy}(13)$ .
Calculate Derivatives: Calculating derivatives: $\newline$ $\frac{d}{dy}(-4y^2) = -8y$ , $\newline$ $\frac{d}{dy}(2y) = 2$ , $\newline$ $\frac{d}{dy}(13) = 0$ . $\newline$ So, $\frac{dx}{dy} = -8y + 2$ .
Substitute $y = 2$ : Now, substitute $y = 2$ into $\frac{dx}{dy}$ to find the slope at the point $(1,2)$ : $\frac{dx}{dy} = -8(2) + 2 = -16 + 2 = -14$ .

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