Question

How do you implicitly differentiate `5y^2=e^(x-4y)-3x`?

Answer 1

`dy/dx= (e^(x-4y)-3)/(10+4e^(x-4y))`

Explanation:

differentiate each part
`5y^2=e^(x-4y)-3x`

`5y^2 => 10ydy/dx`

`e^(x-4y) =>e^(x-4y) -4e^(x-4y)dy/dx`
(this part is the most difficult so I'll do it in a little more depth)

`y=e^(x-4y)` using the rule `y=e^f(x) => f'(x)e^f(x)`

giving us

`y'=(1-4dy/dx)e^(x-4y)`

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

`-3x => -3`

leaving us with

`10y dy/dx=e^(x-4y) -4e^(x-4y)dy/dx-3`

`10y dy/dx+4e^(x-4y)dy/dx=e^(x-4y)-3`

`dy/dx(10y+4e^(x-4y))=e^(x-4y)-3`

`dy/dx=(e^(x-4y)-3)/(10y+4e^(x-4y))`

which is the Answer.