Question

How do you differentiate `xe^y=y-1`?

Answer 1

`(dy)/(dx)=e^y/(1-xe^y)`

Explanation:

Generally differentiation problems involve functions i.e. `y=f(x)` - written explicitly as functions of `x`. However, some functions of `y` are written implicitly as functions of `x`, like the one in question above.

So what we do is to treat `y` as `y=y(x)` and use chain rule. This means differentiating `y` w.r.t. `y`, but as we have to derive w.r.t. `x`, as per chain rule, we multiply it by `(dy)/(dx)`.

Using this, differential of `xe^y=y-1` can be derived as follows:

`x xxd/(dy)*e^yxx(dy)/(dx)+1xxe^y=d/(dy)yxx(dy)/(dx)`

or `xe^y(dy)/(dx)+e^y=(dy)/(dx)`

or `(dy)/(dx)(1-xe^y)=e^y`

or `(dy)/(dx)=e^y/(1-xe^y)`

Answer 2

`e^y/(2-y)`.

Explanation:

#x=e^(-y)(y-1) is from the given solvable-for-x implicit form.

In such problems, use

`x'= (dx)/(dy)=-e^(-y)(y-1)+e^(-y)=(2-y)e^(-y)= 1/(y')`. So,

`(dy)/(dx)=y'=e^y/(2-y)`..