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April 7th, 2019, 02:36 AM   #1
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Weird differentiation result

Hoping some good math people can help me out here. (English is not my native language, so I hope I am using the correct math terms.)

I am asked to differentiate $\displaystyle x*\sqrt{x}$ using the product rule. So letting the first $\displaystyle x$ be f(x) and the $\displaystyle \sqrt{x}$ be g(x)
... the result should be $\displaystyle f'(x)*g(x) + f(x)*g'(x)$
... thus $\displaystyle 1\sqrt{x} + x\frac {1}{2\sqrt{x}}$
(this can be reduced a bit further but that's not the point).

BUT ... Inserting this in GeoGebra (or any other math tool) gives a (for me at least) weird f' result :
$\displaystyle 3*\frac {x}{2\sqrt{x}}$

The plot of both my "manual" product rule derivation and the "weird result" actually gives the exact same curve. Indeed, using Excel to simulate the formulas also confirms that "manual" product rule equation gives the same values and curves as the GeoGebra f' result.

Last edited by skipjack; April 7th, 2019 at 11:20 AM.
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April 7th, 2019, 11:39 AM   #2
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$\displaystyle x = \sqrt{x}\cdot\sqrt{x} \implies \sqrt{x} = \frac{x}{\sqrt{x}}$
$\displaystyle 1\sqrt{x} + x\frac {1}{2\sqrt{x}} = 1\cdot\frac{x}{\sqrt{x}} + x\frac {1}{2\sqrt{x}} = \left(1 + \frac12\right)\frac{x}{\sqrt{x}} = \frac32\frac{x}{\sqrt{x}} = \frac32\sqrt{x}$
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April 7th, 2019, 11:39 AM   #3
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I think you'll feel a bit silly when you see the answer.

$\sqrt{x} + x \dfrac{1}{2\sqrt{x}} = \\

\dfrac{x}{\sqrt{x}} + \dfrac{x}{2\sqrt{x}} = \\

\left(1+\dfrac 1 2\right) \dfrac{x}{\sqrt{x}} = \\

\dfrac 3 2 \dfrac{x}{\sqrt{x}} = 3 \cdot \dfrac{x}{2\sqrt{x}}

The reason you missed it is probably because it's a very goofy use
of radicals. I can't think of a single instance I've expressed

$\sqrt{x} = \dfrac{x}{\sqrt{x}}$

It's precisely the opposite of what is usually done.
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April 7th, 2019, 11:40 AM   #4
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Math Focus: Wibbly wobbly timey-wimey stuff.
You should have simplified it.

$\displaystyle \sqrt{x} + x \cdot \dfrac{1}{2 \sqrt{x}} = \sqrt{x}+ \dfrac{1}{2} \cdot \sqrt{x} = \dfrac{3}{2} \sqrt{x}$


Wow! Busy on the Forum this afternoon!
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April 7th, 2019, 03:21 PM   #5
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Thanks a ton guys - and yes I feel silly. But not because "I forgot" something - more like "I never knew", which might be embarassing. I have simply never seen that $\displaystyle \sqrt{x}=\frac {x}{\sqrt{x}}$. I get it, but ma-an never thought about it. Again thanks guys
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