Very short math question derivatives?

Asked Jun 09 '13 at 17:13

Active Jun 09 '13 at 18:48

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If we have $2x+p+ (2x+p)\cdot \dfrac {dp}{dx} =0$ we express it $\left(\dfrac {dp}{dx} + 1\right) \cdot(2x+p)=0$

What if we have $\;0 =p+ \left[(2x)- \dfrac 1p\right]\cdot \dfrac {dp}{dx}$ how can we express it like the upper form? I am in the middle of a VERY long exercise and I need this please.

edited Jun 09 '13 at 18:48

iostream007

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asked Jun 09 '13 at 17:13

fgf

You can express it in exact form as $(p^2) dx + (2x p -1)dp = 0$ – Amzoti Jun 09 '13 at 17:16
I love you,my virtual soulmate.You complete me.(Literally). – fgf Jun 09 '13 at 17:17
You are very kind, but MSE is OUR virtual soulmate! :-) Regards – Amzoti Jun 09 '13 at 17:18
Ha! That was smart. – fgf Jun 09 '13 at 17:22

Very short math question derivatives?

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