My Math Forum Derivative of x times the absolute value of x

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 June 27th, 2012, 04:39 PM #1 Newbie   Joined: Jun 2012 Posts: 1 Thanks: 0 Derivative of x times the absolute value of x Ok...so what is the derivative of: h(x) = x*|x| at (0,0)
 June 27th, 2012, 06:02 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Derivative of x times the absolute value of x We are given: $h(x)=x|x|=x\sqrt{x^2}$ Using the product, power and chain rules, we find: $h'(x)=x$$\frac{x}{\sqrt{x^2}}$$+\sqrt{x^2}=\fr ac{2x^2}{\sqrt{x^2}}=2\sqrt{x^2}=2|x|$ Hence: $h'(0)=0$
 June 27th, 2012, 09:51 PM #3 Newbie   Joined: May 2012 Posts: 3 Thanks: 0 Re: Derivative of x times the absolute value of x y=xmodx If x is positive modx=x so y=x^2 and derivative=2x If x is negative modx=-x so y= -x^2 and derivative= -2x
 June 28th, 2012, 09:43 PM #4 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Derivative of x times the absolute value of x x|x| behaves exactly like x^2 with the left half reflected about the -x axis interestingly, |x| has no derivative at x=0 but x|x| has derivative everywhere. http://www.wolframalpha.com/input/?i=pl ... Cx%5E2%29+

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# derivative of x*abs(x)

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