Question

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

Answer 1

`:.dy/dx={3-2e^(2x-4y)}/[4{2y-e^(2x-4y)}].`

OR

`.dy/dx={3-6x+2y^2}/{4(2y-3x+y^2}.`

Explanation:

We write the eqn. of the curve as `: 3x-y^2=e^(2x-4y)..........(1)`

Diff.ing both sides w.r.t. x, `d/dx(3x-4y^2)=d/dxe^(2x-4y).`

`:. d/dx3x-d/dx4y^2=e^(2x-4y)*d/dx(2x-4y).`
`:.3-4d/dy(y^2)dy/dx=e^(2x-4y)(d/dx2x-d/dx4y)`......[Chain Rule]
`:.3-4(2y)dy/dx=e^(2x-4y){2-4dy/dx}.`
`:.3-2e^(2x-4y)=8ydy/dx-4e^(2x-4y)dy/dx=4{2y-e^(2x-4y)}dy/dx.`
`:.dy/dx={3-2e^(2x-4y)}/[4{2y-e^(2x-4y)}].`

This is fairly a good answer, but, we can further simplify it, replacing `e^(2x-4y),` by `3x-y^2` [bcz. of `(1)`], as under :-

`.dy/dx={3-2(3x-y^2)}/[4{2y-(3x-y^2)}]={3-6x+2y^2}/{4(2y-3x+y^2}.`