Question

How do you implicitly differentiate `y = (ln x)^(1/2) y-3+y^2/x`?

Answer 1

For this one, just do the regular product rule, essentially xD

Explanation:

Relating to the title, using the product rule to differentiate `(lnx)^(1/2)y` with respect to `x` would look like this:
`d/dx((lnx)^(1/2)y)=d/dx((lnx)^(1/2))y+(lnx)^(1/2)d/dxy`

You still follow what steps you would usually take for a derivative, this time you're just getting some extra `dy/dx` terms.

Differentiating the whole thing,
`y=(lnx)^(1/2)y-3+y^2/x`
`dy/dx=(d/dx((lnx)^(1/2))y+(lnx)^(1/2)d/dxy)-d/dx(3)+(d/dx(1/x)y^2+1/xd/dx(y^2))`

`dy/dx=(1/(2lnx)1/xy+(lnx)^(1/2)dy/dx)-0+(-1/x^2y^2+1/x2yd/dxy)`

`dy/dx=(y/(2xlnx)+(lnx)^(1/2)dy/dx)+(-y^2/x^2+(2y)/xdy/dx)`

`dy/dx=y/(2xlnx)+(lnx)^(1/2)dy/dx-y^2/x^2+(2y)/xdy/dx`

Now that everything has been differentiated, move all the `dy/dx` terms to the left side and separate it through factorization.

`dy/dx-(lnx)^(1/2)dy/dx-(2y)/xdy/dx=y/(2xlnx)-y^2/x^2`

`dy/dx(1-(lnx)^(1/2)-(2y)/x)=y/(2xlnx)-y^2/x^2`

`dy/dx=(y/(2xlnx)-y^2/x^2)/(1-(lnx)^(1/2)-(2y)/x)`

Do some simplifications, in this case factor out a `1/x^2` from the top and bottom and then multiple the top and bottom by `lnx`

`dy/dx=((yx)/(2lnx)-y^2)/(x^2-x^2(lnx)^(1/2)-2xy)`

`dy/dx=(yx-y^2lnx)/(x^2lnx-x^2(lnx)^(1/2)-2xylnx)`

And there you have it. I'm sure someone else can edit this post and simplify it further, but yeah. Essentially, just do regular differentiation and move then it around to solve for `dy/dx`.