Question

What is the implicit derivative of `1= e^(xy)`?

Answer 1

`(dy)/dx=-y/x`

Explanation:

When we differentiate we have to use the chain rule in conjunction with the product rule.

The left had side is a constant `1` so its derivative with respect to `x` is `0`

For the right hand side we use the chain rule and the product rule.

`e^(xy)[y+x(dy)/dx]`

So together we have

`0=e^(xy)[y+x(dy)/dx]`

Distribute `e^(xy)`

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

Isolate term with `(dy)/dx`

`(dy)/dxxe^(xy)=-ye^(xy)`

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

`(dy)/dx=-y/x`

Answer 2

I like the question and the answer can be written `dy/dx = y/x`, but . . .

Explanation:

`1=e^(xy)` implies that `xy=0` which in turn implies that either `x=0` or `y=0`.

The graph of this equation is the pair of axes.

Here is the graph of `1=e^(xy)` using Socratic's graphing utility:

graph{1=e^(xy) [-10, 10, -5, 5]}