Question

How do you implicitly differentiate `11=(x-y)/(e^y-e^x)`?

Answer 1

Multiply both sides of the equation by `e^y - e^x`
Separate terms containing y to the left and x to the right.
Differentiate each term.
Divide both sides by the coefficient of `dy/dx`

Explanation:

Multiply both sides by `e^y - e^x`

`11e^y - 11e^x = x - y`

Add 11e^x + y to both sides:

`y + 11e^y = 11e^x + x`

Differentiate each term with respect to x:

`(d(y))/dx + (d(11e^y))/dx = (d(11e^x))/dx + (d(x))/dx`

The derivative of y is `dy/dx`:

`dy/dx + (d(11e^y))/dx = (d(11e^x))/dx + (d(x))/dx`

The derivative of `11e^y` is `11e^ydy/dx`

`dy/dx + 11e^ydy/dx = (d(11e^x))/dx + (d(x))/dx`

The derivative of `11e^x` is itself.

`dy/dx + 11e^ydy/dx = 11e^x + (d(x))/dx`

The derivative of x is 1:

`dy/dx + 11e^ydy/dx = 11e^x + 1`

Factor dy/dx from the left side:

`dy/dx(1 + 11e^y) = 11e^x + 1`

Divide both sides `(1 + 11e^y)`

`dy/dx = (11e^x + 1)/(1 + 11e^y)`