Question

Use implicit differentiation to find dy dx for x^2 y^3 + 3y^2 − 5x = −10.

(b) Find the equation of the line tangent to the curve in part a) at the point (1, 2).

Answer 1

given that

x^2*y^3 + 3y^2 - 5x = -10

Using Implicit differentiation:

2x*y^3 + x^2*3*y^2*(dy/dx) + 3*2*y*(dy/dx) - 5*1 = 0

dy/dx*(3x^2y^2 + 6y) = 5 - 2xy^3

dy/dx = (5 - 2xy^3)/(3x^2y^2 + 6y)

Part B.

ow slope of tangent line,going through point (1, 2) will be:

m = (dy/dx)_(1, 2)

m = (5 - 2*1*2^3)/(3*1^2*2^2 + 6*2) = -11/24

Now equation of tangent line will be:

(y - y1) = m*(x - x1)

(y - 2) = (-11/24)*(x - 1)

y - 2 = -11*x/24 + 11/24

24y - 48 = -11x + 11

11x + 24y - 59 = 0

y = -11x/24 + 59/24

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