Pre-Calculus Pre-Calculus Math Forum

 April 7th, 2019, 02:36 AM #1 Newbie   Joined: Apr 2019 From: Denmark Posts: 2 Thanks: 0 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. April 7th, 2019, 11:39 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,757 Thanks: 2138 $\displaystyle x = \sqrt{x}\cdot\sqrt{x} \implies \sqrt{x} = \frac{x}{\sqrt{x}}$ $\displaystyle 1\sqrt{x} + x\frac {1}{2\sqrt{x}} = 1\cdot\frac{x}{\sqrt{x}} + x\frac {1}{2\sqrt{x}} = \left(1 + \frac12\right)\frac{x}{\sqrt{x}} = \frac32\frac{x}{\sqrt{x}} = \frac32\sqrt{x}$ Thanks from topsquark and SteenHansen April 7th, 2019, 11:39 AM #3 Senior Member   Joined: Sep 2015 From: USA Posts: 2,468 Thanks: 1342 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. Thanks from topsquark and SteenHansen April 7th, 2019, 11:40 AM #4 Math Team   Joined: May 2013 From: The Astral plane Posts: 2,205 Thanks: 901 Math Focus: Wibbly wobbly timey-wimey stuff. You should have simplified it. $\displaystyle \sqrt{x} + x \cdot \dfrac{1}{2 \sqrt{x}} = \sqrt{x}+ \dfrac{1}{2} \cdot \sqrt{x} = \dfrac{3}{2} \sqrt{x}$ -Dan Wow! Busy on the Forum this afternoon! Thanks from SteenHansen April 7th, 2019, 03:21 PM #5 Newbie   Joined: Apr 2019 From: Denmark Posts: 2 Thanks: 0 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 Tags differentiation, result, weird Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post J Thomas Physics 8 February 27th, 2015 08:34 AM ZardoZ Real Analysis 6 August 11th, 2011 05:06 PM Wissam Number Theory 10 September 25th, 2010 07:55 AM travo Number Theory 2 August 22nd, 2009 09:27 AM

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