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Find the slope of the tangent line to the graph of −5y3+2y2−4=x-5y^3 + 2y^2 - 4 = x at (3,−1)(3,-1).

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Full solution

Q. Find the slope of the tangent line to the graph of −5y3+2y2−4=x-5y^3 + 2y^2 - 4 = x at (3,−1)(3,-1).
  1. Differentiate implicitly: First, we need to find the derivative of the equation with respect to yy to get the slope of the tangent line. Since the equation is given in terms of yy, we differentiate implicitly.\newlineddy(−5y3+2y2−4)=ddx(x)\frac{d}{dy}(-5y^3 + 2y^2 - 4) = \frac{d}{dx}(x)\newline−15y2+4y=1-15y^2 + 4y = 1
  2. Substitute y=−1y = -1: Next, substitute y=−1y = -1 into the derivative to find the slope at that specific point.−15(−1)2+4(−1)=−15+(−4)=−19-15(-1)^2 + 4(-1) = -15 + (-4) = -19

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