# find dy/dx a) 3x^4y^2 + 4xy - y^4 - x^2 - 4y - 2 = 0 b) x^3 cos(1+y^2) = 3y^2 - x^4 +1

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### 1 Answer

a) You should differentiate the function with respect to x using the product rule in some cases such that:

`(3x^4y^2 + 4xy - y^4 - x^2 - 4y - 2) = 0'`

`12x^3y^2 + 6x^4y*(dy)/(dx) + 4y + 4x*(dy)/(dx) - 4y^3*(dy)/(dx) - 2x - 4*(dy)/(dx) = 0`

You need to factor out `(dy)/(dx)` such that:

`(6x^4y + 4x - 4y^3 - 4)*(dy)/(dx) + 12x^3y^2 + 4y - 2x = 0`

`(6x^4y + 4x - 4y^3 - 4)*(dy)/(dx) = 2x - 4y - 12x^3y^2`

`2(3x^4y + 2x - 2y^3 - 2)*(dy)/(dx) = 2(x - 2y - 6x^3y^2) `

`(dy)/(dx) = (x - 2y - 6x^3y^2)/(3x^4y + 2x - 2y^3 - 2)`

**Hence, evaluating `(dy)/(dx) ` for the given expression yields** `(dy)/(dx) = (x - 2y - 6x^3y^2)/(3x^4y + 2x - 2y^3 - 2).`

b)`(x^3 cos(1+y^2))' = (3y^2 - x^4 +1)` '

`3x^2cos(1+y^2) + 2y*x^3*(-sin(1+y^2))*(dy)/(dx) = 6y*(dy)/(dx) - 4x^3`

You need to isolate the terms that contain (dy)/(dx) to the left side such that:

`2y*x^3*(-sin(1+y^2))*(dy)/(dx) - 6y*(dy)/(dx) = - 3x^2cos(1+y^2) - 4x^3`

You need to factor out `(dy)/(dx)` such that:

`(2y*x^3*(-sin(1+y^2)) - 6y)*(dy)/(dx) = - 3x^2cos(1+y^2) - 4x^3`

`(dy)/(dx) = (6y - 2y*x^3*(-sin(1+y^2)) - 6y)/(3x^2cos(1+y^2)+ 4x^3)`

**Hence, evaluating `(dy)/(dx)` for the given expression yields `(dy)/(dx) = (6y - 2y*x^3*(-sin(1+y^2)) - 6y)/(3x^2cos(1+y^2) + 4x^3).` **