Question

Further mathematics: Find the derivative of the function x^2 y + (xy)^3 +3x=0 at (-1 , 3)​

Answer 1

Explanation:

x²y + (xy)³ + 3x = 0

x²y + x³y³ + 3x = 0

Using implicit differentiation and product rule:

2xy + x²(dy/dx) + 3x²y³ + 3y²x³(dy/dx) + 3 = 0

x²(dy/dx) + 3x³y²(dy/dx) = -3x²y³ - 2xy - 3

dy/dx(x²(3xy² + 1)) = -3x²y³ - 2xy - 3

dy/dx = (-3x²y³ - 2xy - 3)/(x²(3xy² + 1))

`{ ((dy)/(dx))}^( - 1) _(3) = \frac{( - 3{( - 1)}^(2) ({3})^(3) - 2( - 1)(3) - 3) }{({( - 1)}^(2)(3( - 1){(3)}^(2) + 1)} \\ \\ { ((dy)/(dx))}^( - 1) _(3) = (( -81 + 6 - 3))/(( - 27 + 1)) \: \\ \\ { ((dy)/(dx))}^( - 1) _(3) = ( - 78)/( - 26) \\ \\ { ((dy)/(dx))}^( - 1) _(3) = 3`

Answer 2

3.

Explanation:

Implicit differentiation:

x^2 y + (xy)^3 + 3x = 0

x^2 y + x^3y^3 + 3x = 0

Using the product rule:

2x* y + x^2*dy/dx + 3x^2 y^3 + x^3* (d(y^3)/dx) + 3 = 0

2xy + x^2 dy/dx + 3x^2 y^3 + x^3* 3y^2 dy/dx + 3 = 0

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

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

At the point (-1, 3).

the derivative = (6 - 81 - 3)/(1 -27)

= -78/-26

= 3.