Question

How do you use implicit differentiation here? the answer should be (5x^4+4x^3y)/(6xy^2-4y^3) but I don't know how to get there

`x^4(x+y)=y^3(3x-y)`

Answer 1

That answer is not correct for this question. Whoever provided the answer made an error in the product rule. (Or they started with a different equation.)

Explanation:

The given equation is equivalent to

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

Differentiating with respect to `x` we will get 6 terms. Three involving `dy/dx` and three not involving the derivative of `y`. Unless some of them combine, `dy/dx` will be the ratio of two trinomials not two binomials.

`5x^4 + 4x^3y + underbrace(x^4 dy/dx)_("this term is missing") = underbrace(3y^3)_("this term is missing") +underbrace(3x*3y^2 dy/dx)_("this term is incorrect") - 4y^3 dy/dx`

`5x^4 + 4x^3y + x^4 dy/dx = 3y^3 +9xy^2 dy/dx - 4y^3 dy/dx`

`5x^4 + 4x^3y - 3y^3 = 9xy^2 dy/dx - 4y^3 dy/dx - x^4 dy/dx`

Answer 2

The correct answer (verified) is as follows:

`dy/dx = (5x^4 + 4x^3y - 3y^3)/(9xy^2 - 4y^3 - x^4)`

Explanation:

Implicit differentiation is just an application of the chain rule. If `y` is a function of `x` ( ie `y=f(x)`) then your differentiate to get `dy/dx` directly. But if y is not an explicit function of `x` (eg `y^2=f(x)`), then you cannot find `dy/dx` directly and instead differemtiate `y^2` with wrt `y` using the chain rule

So we have `x^4(x+y)=y^3(3x-y)`

I would tackle this by first multiplying out:
`x^5+x^4y = 3xy^3-y^4`

We then differentiate everything wrt `x`
`:. d/dx(x^5)+d/dx(x^4y) = d/dx(3xy^3)-d/dx(y^4)`
`:. 5x^4+d/dx(x^4y) = 3d/dx(xy^3)-d/dx(y^4)`

We will now also need to apply the product rule
`d/dx(uv)=u(dv)/dx+v(du)/dx` to get

`5x^4 + { (x^4)(d/dxy) + (y)(d/dxx^4) } = 3{ (x)(d/dxy^3) + (y^3)(d/dxx) } - d/dx(y^4)`

`5x^4 + x^4dy/dx + 4x^3y = 3{ x(d/dxy^3) + y^3 } - d/dx(y^4)`

Now we can't differentiate functions of y wrt x but we can use the chain rule (and this is the implicit differentiation)

`5x^4 + x^4dy/dx + 4x^3y = 3{ xdy/dxd/dy(y^3) + y^3 } - dy/dxd/dy(y^4)`

This subtle change allows s now to differentiate the non-explicit functions of y;

`5x^4 + x^4dy/dx + 4x^3y = 3{ xdy/dx(3y^2) + y^3 } - dy/dx(4y^3)`

`:. 5x^4 + x^4dy/dx + 4x^3y = 9xy^2dy/dx + 3y^3 - 4y^3dy/dx`

we can gather up the `dy/dx` term on the LHS to get;
`5x^4 + 4x^3y - 3y^3 = 9xy^2dy/dx - 4y^3dy/dx - x^4dy/dx`
`:. (9xy^2 - 4y^3 - x^4)dy/dx = 5x^4 + 4x^3y - 3y^3`
`:. dy/dx = (5x^4 + 4x^3y - 3y^3)/(9xy^2 - 4y^3 - x^4)`