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I'm having trouble with differentiating a expression. I do it one way, wolfram alpha does it another. Let me show you what I mean. The original expression is this:

$$\frac{1}{2u^3}$$

I start by moving the variables to the numerator.

$$2u^{-2}$$

Then I use the power rule to get:

$$-4u^{-3}$$

Or: $$\frac{1}{-4u^3}$$

Wolfram alpha takes a different approach and receives a different answer, it starts off by factoring out a one-half:

$$\frac{1}{2} * \frac{1}{u^2}$$

Then it uses the power rule on the second expression:

$$\frac{1}{2} * -2u^{-3}$$

It then combines the terms and removes the $\frac{2}{2}$ to yeild:

$$\frac{-1}{u^3}$$

I'm not one to argue with a computer, but I wasn't aware factoring out ahead of time was /mandatory/ in situations like this. Is it? Or did I do something wrong in one of my steps? Please let me know.

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  • $\begingroup$ $\frac{1}{2u^2} = (2u^2)^{-1} = 2^{-1}\cdot u^{-2}$, you transformed the $2^{-1}$ into a $2$. $\endgroup$ – Daniel Fischer Jul 11 '14 at 19:40
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$$\frac{1}{2u^3}=\frac{1}{2} u^{-3}$$

$$\left(\frac{1}{2u^3}\right)'=\frac{1}{2} (u^{-3})'=\frac{-3}{2}u^{-4}=\frac{-3}{2u^4}$$

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Calculus is easy. It's the Algebra and Geometry (but mostly Algebra) that trip students up usually.

$$\frac{1}{2u^2}=\frac{u^{-2}}{2}\ne 2u^{-2}$$

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your expression is just like: $0.5u^{-3}$ which we derive like every single argument polynomial; $-3 *0.5 *u^{-3-1}$ which is in fact just: $-1.5 * u^{-4}$

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