Question

How do you implicitly differentiate `2=(x+y)^2-y^2-x+e^(3x+y^2)`?

Answer 1

`dy/dx = (-2x - 2y + 1 - 3e^(3x+y^2))/(2x + 2ye^(3x+y^2))`

Explanation:

For this, we take derivatives of all terms with respect to `x`. When differentiating a `y` term, we must note the presence of `dy/dx` (often written as `y'` for convenience). After that is done, we solve for `y'`:

`0 = 2(x + y)(1 + 1*y') - 2yy' - 1 + e^(3x+y^2)*(3+2yy')`

`0 = (2x + 2y)(1 + y') - 2yy' - 1 + 3e^(3x+y^2) + 2yy'e^(3x+y^2)`

`0 = 2x + 2xy' + 2y + 2yy' - 2yy' - 1 + 3e^(3x+y^2) + 2yy'e^(3x+y^2)`

`0 = 2x + 2xy' + 2y - 1 + 3e^(3x+y^2) + 2yy'e^(3x+y^2)`

`-2x - 2y + 1 - 3e^(3x+y^2) = 2xy' + 2yy'e^(3x+y^2)`

`-2x - 2y + 1 - 3e^(3x+y^2) = y'(2x + 2ye^(3x+y^2))`

`(-2x - 2y + 1 - 3e^(3x+y^2))/(2x + 2ye^(3x+y^2)) = y'`