Question

How can I find the derivatives of the implicitly set function y=y(x), set with the equation `xe^y + ye^x - e^(xy) = 0`? the answer I have is y' = ye^xy -e^y -ye^x/xe^y + e^x -xe^xy

Answer 1

`dy/dx = (ye^(xy)-ye^x-e^y)/(xe^y + e^x -xe^(xy))`

Explanation:

Given: `xe^y + ye^x - e^(xy) = 0`

Differentiate each term with respect to x:

`(d(xe^y))/dx + (d(ye^x))/dx -(d(e^(xy)))/dx = (d(0))/dx" [1]"`

The first term requires the use of the product rule:

`(d(uv))/dx = (du)/dxv+u(dv)/dx`

Where `u = x and v = e^y`, then `(du)/dx =1 and (dv)/dx = (d(e^y))/dx`

We need the chain rule for `(d(e^y))/dx`:

`(d(e^y))/dx = (d(e^y))/dydy/dx`

`(d(e^y))/dx = e^ydy/dx`

Substituting into the product rule:

`(d(xe^y))/dx = e^y+xe^ydy/dx" [2]"`

Substitute equation [2] into equation [1]:

`e^y+xe^ydy/dx + (d(ye^x))/dx -(d(e^(xy)))/dx = (d(0))/dx" [1.1]"`

The next term, also, requires the use of the product rule:

`(d(uv))/dx = (du)/dxv+u(dv)/dx`

Where `u = y, and v = e^x`, then `(du)/dx = dy/dx and`(dv)/dx = e^x#

Substituting into the product rule:

`(d(ye^x))/dx = dy/dxe^x+ye^x" [3]"`

Substitute equation [3] into equation [1.1]:

`e^y+xe^ydy/dx + dy/dxe^x+ye^x -(d(e^(xy)))/dx = (d(0))/dx" [1.2]"`

If we let `u = xy`, then we can use the chain rule for the next term:

`(d(e^u))/dx = (d(e^u))/(du)(du)/dx`

`(d(e^u))/dx = e^u(du)/dx`

But we shall need the product rule to compute `(du)/dx`

`(du)/dx = y +xdy/dx`

`(d(e^u))/dx = e^u(y +xdy/dx)`

Reverse the substitution for u:

`(d(e^(xy)))/dx = e^(xy)(y +xdy/dx)`

Use the distributive property:

`(d(e^(xy)))/dx = ye^(xy) +xe^(xy)dy/dx" [4]"`

Substitute equation [4] into equation [1.2] (remember to distribute the leading -1):

`e^y+xe^ydy/dx + dy/dxe^x+ye^x -ye^(xy) -xe^(xy)dy/dx = (d(0))/dx`

The derivative of 0 is 0:

`e^y+xe^ydy/dx + dy/dxe^x+ye^x -ye^(xy) -xe^(xy)dy/dx = 0`

Move all of the terms that do NOT contain `dy/dx` to the right:

`xe^ydy/dx + dy/dxe^x -xe^(xy)dy/dx = ye^(xy)-ye^x-e^y`

Factor out `dy/dx` on the left:

`(xe^y + e^x -xe^(xy))dy/dx = ye^(xy)-ye^x-e^y`

Divide both sides by `(xe^y + e^x -xe^(xy))`:

`dy/dx = (ye^(xy)-ye^x-e^y)/(xe^y + e^x -xe^(xy))`

Answer 2

`y'=-1/x`

Explanation:

notice that `ye^x` and `e^xy` are same.Therefore `ye^x-e^xy=0`

this will give you `xe^y=0`

applying product rule `(f.g)'=f.g'+f'.g`

so `y'xe^y+e^y=0`
rearrange the equation,
`y'xe^y=-e^y`
`y'=-e^y/(xe^y)`
y'=`-1/x`

alternatively,
`ln(xe^y)=0`
`lnx +ln e^y=0`
`lnx+y=0`
`y=-ln x`
`y'=-1/x`