I don't seem to understand how to completely differentiate this equation on both sides (implicit differentiation):

x^3-xy+y^2=4

Am I doing this right:

3x^2-(x'y+xy')+2yy'=0

Any help would be appreciated

Thanks

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- Oct 21st 2008, 09:11 PMPMoNEY23[SOLVED] Implicit Differentiation
I don't seem to understand how to completely differentiate this equation on both sides (implicit differentiation):

x^3-xy+y^2=4

Am I doing this right:

3x^2-(x'y+xy')+2yy'=0

Any help would be appreciated

Thanks - Oct 21st 2008, 09:19 PMU-God
Yes it looks like you're on the right track (assuming you're differentiating with respect to x).

You've written the product rule in your parenthesis, but you are able to actually differentiate the x parts directly such that your answer would look like:

$\displaystyle 3x^2 -y - x\frac{dy}{dx}+2y\frac{dy}{dx} = 0 $ - Oct 21st 2008, 09:32 PMPMoNEY23
But I don't know what you mean by differentiating by themselves or in parts, for example where did that y come from and why isn't it y'?

I don't know if that made sense? - Oct 21st 2008, 09:48 PMU-God
Okay so you know that the product rule is: $\displaystyle \frac{d}{dx} [fg] = \frac{df}{dx}g + \frac{dg}{dx}f $ assuming both f and g are functions of x.

however, if f(x) = x, then f'(x) = 1

Therefore $\displaystyle \frac{df}{dx}g = g $

In your case, g = y.

When I said you can differentiate the x parts directly, I was referring to how inside the parenthesis you left the derivative of x as x'. You could have omitted this and left it as 1. - Oct 21st 2008, 09:54 PMPMoNEY23
I see that makes much more sense, well I got that one down, THanks