Given the curve xy^2 + yx^3 +x + 51= 0 , find an equation of the tangent line at the point (3,-3). Please write your answer in point slope form. I'm just having a problem deriving this with respect to y. Can you please show work, Thanks!
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I think if you use implicit derivative, and the product rule, and the chain rule for y²:
So derivative of xy² = (1)y² + x(2y)(dy/dx). Then derivative of yx³ = (dy/dx)(x³) + y(3x²)
derivative of x = 1. derivative of 51 = 0.
So now we have (1)y² + x(2y)(dy/dx) + (dy/dx)(x³) + y(3x²) + 1 = 0.
Plug in x = 3, and y = -3, and isolate dy/dx: (1)(-3)² + (3)(2*(-3))(dy/dx) + (dy/dx)(3^3) + (-3)(3*3^2) + 1 = 0.
dy/dx = 71/9. This is the slope of the line at the point (3,-3). Use point slope form: (y - y1) = m(x - x1).
y - -3 = (71/9)(x - 3). y = (71/9)x - 71/3 - 3 :
y = (71/9)x - 80/3
So derivative of xy² = (1)y² + x(2y)(dy/dx). Then derivative of yx³ = (dy/dx)(x³) + y(3x²)
derivative of x = 1. derivative of 51 = 0.
So now we have (1)y² + x(2y)(dy/dx) + (dy/dx)(x³) + y(3x²) + 1 = 0.
Plug in x = 3, and y = -3, and isolate dy/dx: (1)(-3)² + (3)(2*(-3))(dy/dx) + (dy/dx)(3^3) + (-3)(3*3^2) + 1 = 0.
dy/dx = 71/9. This is the slope of the line at the point (3,-3). Use point slope form: (y - y1) = m(x - x1).
y - -3 = (71/9)(x - 3). y = (71/9)x - 71/3 - 3 :
y = (71/9)x - 80/3