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Use implicit differentiation to find y' and then evaluate y' at (-3,5). 6xy + y + 85=0 y=0 Y'(-3,5) = (Simplify your answer.) ww.

User Shyam K
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After differentiation and evaluating y' at (-3,5). 6xy + y + 85=0 y=0 we got y'(-3, 5) equal to 30/17

Implicit differentiation is a technique of finding the derivative of an equation in which the dependent variable and independent variable are not clearly defined and cannot be solved for the dependent variable directly. Here, we are to use implicit differentiation to find y' and evaluate it at (-3,5).

Let us consider the given equation;6xy + y + 85=0Taking the derivative with respect to x on both sides, we have:$$\frac{d}{dx}\left(6xy + y + 85\right) = \frac{d}{dx} 0$$$$6x\frac{dy}{dx} + 6y + \frac{dy}{dx} = 0$$

Factoring out dy/dx, we have;$$\frac{dy}{dx}(6x + 1) = -6y$$$$\frac{dy}{dx} = \frac{-6y}{6x + 1}$$To find y' at (-3, 5), we will substitute x = -3 and y = 5 into the expression we obtained for y'.Thus, we have;$$y'(-3, 5) = \frac{-6(5)}{6(-3) + 1}$$$$y'(-3, 5) = \frac{-30}{-17}$$$$y'(-3, 5) = \frac{30}{17}$$Therefore, y'(-3, 5) = 30/17.I hope this helps.

User Denis Palnitsky
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