Question

How do you differentiate `e^(xy)+y=x-1`?

Answer 1

Differentiate each term, put each term back into the equation, and then solve for `dy/dx`

Explanation:

Differentiate: `e^(xy)+y=x-1`

The first term:

Let `u = xy`, then `(du)/dx = (d(xy))/dx`

Use the product rule:

`(d(xy))/dx = (d(x))/dxy+x(d(y))/dx`

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

Use the chain rule:

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

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

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

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

`(d(e^(xy)))/dx = ye^(xy) + xe^(xy)dy/dx`

The second term:

`(d(y))/dx = dy/dx`

The third term:

`(d(x))/dx = 1`

The fouth term:

`(d(-1))/dx = 0`

Put the terms back into the equation:

`ye^(xy) + xe^(xy)dy/dx + dy/dx = 1`

Solve for `dy/dx`:

`xe^(xy)dy/dx + dy/dx = 1- ye^(xy)`

`(xe^(xy) + 1)dy/dx = 1- ye^(xy)`

`dy/dx = (1- ye^(xy))/(xe^(xy) + 1)`

Answer 2

See explanation

Explanation:

First, we differentiate both sides with respect to x:

`d/dx e^(xy) + dy/dx = d/dx(x+1)`
`->d/dxe^(xy)+ y' = 1` (1)

For differentiating `e^(xy)`, recall that `d/dx e^u = (du)/dx *e^u`

With `u=xy, (du)/dx = d/dx (xy)`

Using the chain rule...

`d/dx (xy) = dx/dx*y + x*dy/dx = y + xy'`

Then back to (1), we get:
`d/dxe^(xy) + y' = 1 -> ye^(xy)+y'xe^(xy) + y' = 1`

We now move all terms without y' as a factor to the right hand side...

`->y'xe^(xy) + y' = 1 -ye^(xy)`

Factor out y' from the left hand side...

`-> y' (xe^(xy)+1)= 1 -ye^(xy)`

And divide both sides by `(xe^(xy)+1)..`

`y' = (1-ye^(xy))/(xe^(xy)+1`