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Find the slope of the tangent line to the graph of x=y3+8y259x = -y^3 + 8y^2 - 59 at (5,4)(5,4).

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Q. Find the slope of the tangent line to the graph of x=y3+8y259x = -y^3 + 8y^2 - 59 at (5,4)(5,4).
  1. Find Derivative of xx: First, we need to find the derivative of xx with respect to yy to get the slope of the tangent line. Since x=y3+8y259x = -y^3 + 8y^2 - 59, differentiate each term with respect to yy.dxdy=3y2+16y\frac{dx}{dy} = -3y^2 + 16y
  2. Substitute y=4y = 4: Next, substitute y=4y = 4 into the derivative to find the slope at that specific point.\newlinedxdy\frac{dx}{dy} at y=4y = 4 = 3(4)2+16(4)=48+64=16-3(4)^2 + 16(4) = -48 + 64 = 16

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