Question

Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy. x5y2 − x4y + 2xy3 = 0

Answer 1

dx/dy = (x^4 - 2x^5y - 6xy^2) / (5x^4y^2 - 4x^3y + 2y^3).

Explanation:

x^5y^2 − x^4y + 2xy^3 = 0

Applying the Product and Chain Rules:

y^2*5x^4*dx/dy + 2y*x^5 - (y*4x^3*dx/dy + x^4) + (y^3* 2*dx/dy + 3y^2*2x) =0

Separating the terms with derivatives:

y^2*5x^4*dx/dy - y*4x^3*dx/dy + y^3* 2*dx/dy = x^4 - 2y*x^5 - 3y^2*2x

dx/dy = (x^4 - 2x^5y - 6xy^2) / (5x^4y^2 - 4x^3y + 2y^3)

Answer 2

`(dx)/(dy)=(-2x^5y+x^4-6xy^2)/(5x^4y^2-4x^3y+2y^3)`

Explanation:

The given equation is

`x^5y^2-x^4y+2xy^3=0`

Differentiate with respect to y.

`(d)/(dy)(x^5y^2)-(d)/(dy)(x^4y)+(d)/(dy)(2xy^3)=0`

Using product rule we get

`x^5(d)/(dy)(y^2)+y^2(d)/(dy)(x^5)-x^4(d)/(dy)(y)-y(d)/(dy)(x^4)+2x(d)/(dy)(y^3)+2y^3(d)/(dy)(x)=0`
`(fg)'=fg'+gf'`

`x^5(2y)+y^2(5x^4(dx)/(dy))-x^4(1)-y(4x^3(dx)/(dy))+2x(3y^2)+2y^3(dx)/(dy)=0`

`2x^5y+5x^4y^2(dx)/(dy)-x^4-4x^3y(dx)/(dy)+6xy^2+2y^3(dx)/(dy)=0`

Isolate
`(dx)/(dy)` terms on left side.

`5x^4y^2(dx)/(dy)-4x^3y(dx)/(dy)+2y^3(dx)/(dy)=-2x^5y+x^4-6xy^2`

`(5x^4y^2-4x^3y+2y^3)(dx)/(dy)=-2x^5y+x^4-6xy^2`

Isolate
`(dx)/(dy)` term.

`(dx)/(dy)=(-2x^5y+x^4-6xy^2)/(5x^4y^2-4x^3y+2y^3)`

Therefore the value of
`(dx)/(dy)` is
`(-2x^5y+x^4-6xy^2)/(5x^4y^2-4x^3y+2y^3)`.