Question

How do you differentiate `e^x/y=4x-y`?

Answer 1

`(e^x - 4y)/(4x - 2y) = dy/dx`

Explanation:

From `e^(x)/y = 4x - y` we have that

`e^x = 4xy - y^2`

So differentiation both sides,

`d/dx(e^x) = d/dx(4xy) - d/dx(y^2)`

Using the product rule

`e^x = 4xd/dx(y) + yd/dx(4x) - d/dx(y^2)`

`e^x = 4xd/dx(y) + 4y - d/dx(y^2)`

Using the chain rule `d/dx = d/dy*dy/dx`

`e^x = 4x*d/dy(y)*dy/dx + 4y - d/dy(y^2)*dy/dx`

`e^x = 4x*dy/dx + 4y - 2y*dy/dx`

`e^x - 4y = (4x - 2y)*dy/dx`

`(e^x - 4y)/(4x - 2y) = dy/dx`

However, as `e^x = 4xy - y^2`, you might want to say it as

`(4xy - y^2 - 4y)/(4x - 2y) = dy/dx`