Implicit Differentiation Watch

Sazi16
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#1
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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
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Kvothe the Arcane
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(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.

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Sazi16
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(Original post by keromedic)
I think you've differentiated improperly.

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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
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Kvothe the Arcane
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(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+y^2, then y \times \dfrac{d}{dx}(e^{-2x})+e^{-2x} \times \dfrac{d}{dx}y=2+\dfrac{d}{dx}y^  2
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Sazi16
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So from what your saying would it be for the RHS simply 2 +2y?
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theDanIdentity
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(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+y^2, then y \times \dfrac{d}{dx}(e^{-2x})+e^{-2x} \times \dfrac{d}{dx}y=2+\dfrac{d}{dx}y^  2

\boxed{x=y^n \\ 1=ny^{n-1} \dfrac{dy}{dx}}

hey, how on earth did you write in 'equation'..? o.0
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Kvothe the Arcane
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(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.
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