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[(f(x+h) - f(x))/h] versus [(f(x) - f(0))/x]

  1. Dec 6, 2011 #1
    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. jcsd
  3. Dec 6, 2011 #2


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    Let [itex]x-x_0 = h[/itex]. Your second formula then becomes your first.
  4. Dec 6, 2011 #3
    So they are the same thing? And should produce the same results?
  5. Dec 6, 2011 #4


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    This is correct. Replacing the general x with x = a, gives us:

    [tex]lim_{h \to 0} \frac{f(a+h)-f(a)}{h}[/tex]

    If this limit exists, the derivative is defined for x = a.

    [itex]\lim_{x \to a} \frac{f(x)-f(a)}{x-a}[/itex] is equivalent to [itex]lim_{h \to 0} \frac{f(a+h)-f(a)}{h}[/itex]. In the first case, you're basically taking the limit of the slope of lines that get closer and closer to a, but both limits are equivalent. If the limit exists, the derivative is defined at x = a.
  6. Dec 6, 2011 #5
    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?


    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?
    Last edited: Dec 6, 2011
  7. Dec 6, 2011 #6


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    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)= x1/2- 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
    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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