
#1
Dec611, 06:15 PM

P: 300

I have this problem, f(x) = (√x)  2x On some interval [0,1].
One of the things im supposed to do is to show that it's differentiable on the interval. Now, I am aware that the definition of the derivative is f'(x) = lim as h[itex]\rightarrow[/itex] 0 of [itex]\frac{f(x+h)  f(x)}{h}[/itex] and that this formula is used to produce the derivative function of f(x). And that replacing the x with a known x value into the formal definition should produce a derivative value at given x value.. so that using this definition of the derivative would tell me whether or not a function at a single x value is differentiable. First Question, is what I said thus far correct? Given what I said above is correct, I don't want to find the derivative of f(x) = (√x)  2x at any one single value. I want to find whether or not it is differentiable on the interval [0,1]. How does the formula lim as x→[itex]x_{0}[/itex] of f(x)  f([itex]x_{0}[/itex])/ x  [itex]x_{0}[/itex] come into play and how does it relate to the above definition of a derivative formula? 



#2
Dec611, 06:23 PM

HW Helper
P: 1,391

Let [itex]xx_0 = h[/itex]. Your second formula then becomes your first.




#3
Dec611, 06:25 PM

P: 300

So they are the same thing? And should produce the same results?




#4
Dec611, 06:27 PM

HW Helper
P: 805

[(f(x+h)  f(x))/h] versus [(f(x)  f(0))/x][tex]lim_{h \to 0} \frac{f(a+h)f(a)}{h}[/tex] If this limit exists, the derivative is defined for x = a. 



#5
Dec611, 06:45 PM

P: 300

So to find where f(x)= √x  2x is differentiable , given the interval [0,1], what do I do?
I could say 2x is differentiable everywhere because I know its graph is a line. And √x exists at x>0 but how can I confirm its differentiability? Do I set up f'x = lim h→0 of ((√(x+h)  2(x+h))  (√(x)  2x))/h and evaluate the formula to come up with the general derivative function for the original function? Or f'x = lim x→a of ((√(x)  2x)  (√(a)  2a))/ x  a and altervatively evaluate this version to come up with a general derivative function for the original function? or should I make "x" value from the first formula equal some value from the interval...or let "a" equal some specific value? Am I supposed to know where its differentiable already, and then simply choose that undifferentiable value into the formula to prove its undifferentiability? 



#6
Dec611, 08:25 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

In order to determine whether or not a function is differentiable on a set, look at the generic formula for the derivative and look for places where that fails.
Here, f(x)= x^{1/2} x. The standard formula for its derivative at [itex]x= x_0[/itex] is [tex]\lim_{h\to 0}\frac{(x_0+ h)^{1/2} (x_0+h) x_0^{1/2}+ x_0}{h}= \lim_{h\to 0}\frac{(x_0+ h)^{1/2} x_0^{1/2} (x_0+ h x_0)}{h}[/tex] [tex]= \lim_{h\to 0}\frac{(x_0+h)^{1/2} x_0^{1/2} h}{h}[/tex] It should be clear that the "h" on the end will give 1 so we need to focus on the square root. Rationalize the numerator by multiplying both numerator and denominator by [itex](x_0)^{1/2}+ x_0^{1/2}[/itex] to get [tex]\frac{x_0+ h x_0}{h((x_0+h)^{1/2}+ x_0^{1/2})}[/tex][tex]= \frac{h}{h((x_0+ h)^{1/2}+ x_0^{1/2})}[/tex][tex][/tex][tex]= \frac{1}{(x_0+ h)^{1/2}+ x_0^{1/2}}[/tex] Now that the h is gone from the denominator, we can take the limit by setting h equal to 0. That gives [tex]\frac{1}{2\sqrt{x_0}}[/tex] As long as [itex]x_0\ne 0[/itex]. So this is differentiable every where except at x= 0. If you already know the derivative laws and are not required to use the defining formula (which I did only because you mentioned it specifically) you could argue that the derivative of [itex]\sqrt{x}= x^{1/2}[/itex] is [itex](1/2)x^{1/2 1}= (1/2)x^{1/2}[/itex] and the derivative of x is, of course, 1, so the derivative of [itex]\sqrt{x} x[/itex] is [itex](1/2)x^{1/2} 1[/itex] which is defined for all x except x= 0. 


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