The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Implicit Differentiation

Hi so I have an equation which I am asked to differentiate ye^-2x=2x+y^2
I know the first bit is a product (I think that's what it's called) where I have to make it uv then use udv + vdu but then I have something that looks like this:

(-2ye^-2x+e^-2x)Dy/dx=2dx + 2ydy

And I don't know how to amalgamate the dys and dxs ... I hope this makes sense... Thanx
Original post by Sazi16
Hi so I have an equation which I am asked to differentiate ye^-2x=2x+y^2
I know the first bit is a product (I think that's what it's called) where I have to make it uv then use udv + vdu but then I have something that looks like this:

(-2ye^-2x+e^-2x)Dy/dx=2dx + 2ydy

And I don't know how to amalgamate the dys and dxs ... I hope this makes sense... Thanx

I think you've differentiated improperly.

Working?
Reply 2
Avatar for Sazi16
Sazi16
OP
Original post by keromedic
I think you've differentiated improperly.

Working?


For the first half of the equation I have (y*e^-2x*-2+e^-2x*1)which is the dy/dx of ye^-2x then for the second half I have 2dx + 2ydy differentiating each term with respect to x/y... It's just that now I have one part of the equation with dy/dx and the other separate dx and dy
Original post by Sazi16
For the first half of the equation I have (y*e^-2x*-2+e^-2x*1)which is the dy/dx of ye^-2x then for the second half I have 2dx + 2ydy differentiating each term with respect to x/y... It's just that now I have one part of the equation with dy/dx and the other separate dx and dy

I see...well you're close with the LHS. You have the product rule and implicit differentiation. But dy/dx is not supposed to be multiplied by each term. On the RHS, I'm not sure how you have what you have. When you differentiate a function of x with respect to x, you don't have a dy/dx.

It might be helpful to review the topics.

But I'll start you of
If ye−2x=2x+y2ye^{-2x}=2x+y^2, then y×ddx(e−2x)+e−2x×ddxy=2+ddxy2y \times \dfrac{d}{dx}(e^{-2x})+e^{-2x} \times \dfrac{d}{dx}y=2+\dfrac{d}{dx}y^2
(edited 9 years ago)
Reply 4
Avatar for Sazi16
Sazi16
OP
So from what your saying would it be for the RHS simply 2 +2y?
Original post by keromedic
I see...well you're close with the LHS. You have the product rule and implicit differentiation. But dy/dx is not supposed to be multiplied by each term. On the RHS, I'm not sure how you have what you have. When you differentiate a function of x with respect to x, you don't have a dy/dx.

It might be helpful to review the topics.

But I'll start you of
If ye−2x=2x+y2ye^{-2x}=2x+y^2, then y×ddx(e−2x)+e−2x×ddxy=2+ddxy2y \times \dfrac{d}{dx}(e^{-2x})+e^{-2x} \times \dfrac{d}{dx}y=2+\dfrac{d}{dx}y^2

x=yn1=nyn−1dydx\boxed{x=y^n \\ 1=ny^{n-1} \dfrac{dy}{dx}}



hey, how on earth did you write in 'equation'..? o.0
Original post by Sazi16
So from what your saying would it be for the RHS simply 2 +2y?

No, read this.
Original post by theDanIdentity
hey, how on earth did you write in 'equation'..? o.0


Sorry, I was still editing. I couldn't figure out how to insert a new line in the box so deleted it.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you