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Homework Help: Help with derivative of

  1. Sep 30, 2004 #1
    Derivative of square root:
    1) Use derivative rule (f(x) = f(x+h) - f(x) / h) to find square root of (6x-4)

    I got: 6 / ((square root of 6x+6h-4) + (square root of 6x-4))
    but is this in the simpliest form? I thought you can't have a square root in the denominator.?.

    2) (derivative) 3 / (x+1)

    I got 1 but the I got it wrong and I tried redoing it but still got it wrong. Can someone please help?
  2. jcsd
  3. Sep 30, 2004 #2


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    Maybe it's just me, but this question doesn't seem to make any sense at all.

    First, check for the answer doing the easy way.
    You're given this:

    [tex]3(x + 1)^{-1}[/tex]

    so the derivative is this:

    [tex]3(-1)(x + 1)^{-2}(1)[/tex]

    [tex]\frac{-3}{x^2 + 2x + 1}[/tex]

    Now to actually do the problem the way you're supposed to, it's just a matter of add/subtracting fractions

    [tex]\frac{\frac{3}{x + h + 1} - \frac{3}{x + 1}}{h}[/tex]

    [tex]\frac{3(x + 1) - 3(x + h + 1)}{(x + h + 1)(x + 1)(h)}[/tex]

    If you keep working at it, you eventually will get the right answer :wink:
    (and yes, I actually did get to the answer working from the formula above)
    Last edited: Sep 30, 2004
  4. Sep 30, 2004 #3


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    Perhaps you meant find the derivative of square root of (6x-4)?

    Your main problem is that in "6/((square root of 6x+6h-4) + (square root of 6x-4))" you've left out an important part: the limit as h-> 0.

    if f(x)= √(6x-4), then f(x+h)= √(6x+ 6h-4) and so the "difference quotient" is [itex] \frac{\sqrt{6x+6h-4}-\sqrt{6x-4}}{h}[/itex]. Since you want to be able to cancel that h in the denominator try "rationalizing the numerator" by multiplying numerator and denominator by [itex]\sqrt{6x+6h-4}+\sqrt{6x-4}[/itex]. In the numerator we get the "difference of two squares", 6x+6h-4- 6x+4= 6h. In the denominator we have the complicated [itex]h\sqrt{6x+6h-4}+\sqrt{6x-4}[/itex] but that's okay: the difference quotient is now [itex]\frac{6h}{h\sqrt{6x+6h-4}+\sqrt{6x-4}}[/itex] and we can cancel the "h"s to get [itex]\frac{6}{\sqrt{6x+6h-4}+\sqrt{6x-4}}[/itex]. Now take the limit as h-> 0 to get [itex]\frac{6}{2\sqrt{6x-4}}[/itex].

    f(x)= 3/(x+1) so f(x+h)= 3/(x+h+1) and the difference quotient is [itex]\frac{\frac{3}{x+h+1}-\frac{3}{x+1}}{h}[/itex]. To subtract those fractions in the numerator, you have to get the common denominator: x+1 and x so we have [itex]\frac{\frac{x+1- (x+h+1}{x(x+h+1}}{h}[/itex]
    That gives [itex]\frac{-h}{hx(x+h+1}= \frac{-1}{x(x+h+1)}[/itex]
    Now take the limit as h goes to 0.
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