December 12th, 2015, 02:19 PM  #1 
Member Joined: May 2015 From: Earth Posts: 64 Thanks: 0  Limit rules trick question?
Use the definition of the derivative (and limit rules) to find the derivative of x^3 at x = 0 Is this a trick question? Like just from looking at it, I know the derivative of an absolute value such as x^3 does not equate at x=0, right? 
December 12th, 2015, 02:58 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 
$\displaystyle f'(a) = \lim_{x \to a} \frac{f(x)  f(a)}{x  a}$ $\displaystyle f'(0) = \lim_{x \to 0} \frac{x^3  0}{x  0}$ $x < 0 \implies x^3 = x^3$ $\displaystyle \lim_{x \to 0^} \frac{x^3}{x}$ $\displaystyle \lim_{x \to 0^} x^2 = 0$ $x \ge 0 \implies x^3 = x^3$ $\displaystyle \lim_{x \to 0^+} \frac{x^3}{x}$ $\displaystyle \lim_{x \to 0^+} x^2 = 0$ limit from the left = limit from the right, therefore ... $\displaystyle f'(0) = \lim_{x \to 0} \frac{x^3  0}{x  0} = 0$ 
December 12th, 2015, 04:53 PM  #3 
Member Joined: May 2015 From: Earth Posts: 64 Thanks: 0 
So you're saying apply the sandwich rule or whatever? To take the derivative on both sides to get the limit? Upon looking at the graph I see there is no edge there like I thought there was. If there was an edge, there would be no limit there, or derivative? How does that all I work? I forget. Thanks. 
December 12th, 2015, 06:23 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,926 Thanks: 2205 
It's straightforward to show from first principles that ax³ has derivative 3ax², with value 3a*0 = 0 at x = 0. Letting a = sgn(x) doesn't affect this result. That wouldn't work for x, as the derivatives of x and x take different values at x = 0. The derivative of x³ is 3xx, as can be verified by differentiating (x²)$^{3/2}$ using the usual rules. 

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