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Find the slope of the tangent line to the graph of x=7y3+y2+10x = 7y^3 + y^2 + 10 at (4,−1)(4,-1).

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Full solution

Q. Find the slope of the tangent line to the graph of x=7y3+y2+10x = 7y^3 + y^2 + 10 at (4,−1)(4,-1).
  1. Find Derivative of xx: First, we need to find the derivative of xx with respect to yy, since the equation is given in terms of yy. We differentiate each term separately.\newlinedxdy=21y2+2y\frac{dx}{dy} = 21y^2 + 2y
  2. Substitute y=−1y = -1: Next, we substitute y=−1y = -1 into the derivative to find the slope of the tangent line at that point.\newlinedxdy=21(−1)2+2(−1)=21−2=19\frac{dx}{dy} = 21(-1)^2 + 2(-1) = 21 - 2 = 19

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