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Derivative of a Function

  1. Mar 5, 2012 #1
    1. The problem statement, all variables and given/known data

    Get the first derivative of [itex]f(x)=\sqrt{5x-4}[/itex]


    2. Relevant equations



    3. The attempt at a solution

    The answer given by Wolfram Alpha is [itex]f'(x)=\frac{5}{2\sqrt{5x-4}}[/itex].

    The confusion I'm having is what method I should use. Wolfram Alpha suggests that I use the Chain Rule. I want to know if it is possible to use Power Rule since [itex]f(x)=\sqrt{5x-4}[/itex] can be written as [itex]f(x)=(5x-4)^{1/2}[/itex]. However, the answer wouldn't be the same as the answer given by Wolfram. I got [itex]f'(x)=\frac{1}{2\sqrt{5x-4}}[/itex]

    I also tried doing the long way [itex]\frac{f(x+h) - f(x)}{h}[/itex] and couldn't proceed since I can't figure out how to simplify it.

    Capture.JPG

    Help would be very much appreciated!!
     
    Last edited: Mar 5, 2012
  2. jcsd
  3. Mar 5, 2012 #2
    Can you please illustrate how you went through the power rule?
     
  4. Mar 5, 2012 #3
    Here :)

    [itex]f'(x)= (5x-4)^{1/2}[/itex]

    Let [itex](y)= 5x-4[/itex]

    [itex]f'(x)= y^{1/2}[/itex]
    [itex]f'(x)= (\frac{1}{2})y^{\frac{1}{2} - 1}[/itex]
    [itex]f'(x)= \frac{y^{-1/2}}{2}[/itex] --- bring y^(-1/2) to make the exponent positive
    [itex]f'(x)= \frac{1}{2(y^{1/2})}[/itex]
    [itex]f'(x)= \frac{1}{2\sqrt{y}}[/itex]

    Then, since [itex](y)= 5x-4[/itex], then

    [itex]f'(x)= (\frac{1}{2\sqrt{5x-4}})[/itex]

    However, I don't think that this is mathematically correct. But, this is exactly how my teacher explained it to me when I asked her. Thanks in advance for helping!! :)
     
  5. Mar 5, 2012 #4

    Mark44

    Staff: Mentor

    You need to use the chain rule as well (the chain rule form of the power rule).

    You have f(x) = (u)1/2, with u = 5x - 4, so f'(x) = (1/2)u-1/2 * du/dx. What you're missing is the du/dx factor.
     
  6. Mar 5, 2012 #5
    The thing is that the power rule doesn't work on composite functions. I see that they have already helped you in previous posts but Ill give you a suggestion. Whenever you encounter a function that can be differentiated with the power rule, differentiate it with the chain rule instead and you'll get a better sense of it. That's what I did.
     
  7. Mar 7, 2012 #6
    Thanks for the replies!!! :)

    So in short, you cannot JUST use the Power Rule? You really have to use the chain rule?
    And also, is it possible to use the long method? Can someone please show me or at least guide me to solve it?? :)
     
  8. Mar 7, 2012 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    [tex]f(x+h)- f(x)= \sqrt{5(x+h)- 4}- \sqrt{5x- 4}[/tex]
    "rationalize" the numerator- multiply by
    [tex]\frac{\sqrt{5(x+h)- 4}+ \sqrt{5x-4}}{\sqrt{5(x+h)- 4}+ \sqrt{5x-4}}[/tex]
    to get
    [tex]\frac{5(x+h)-4 -(5x-4)}{\sqrt{5(x+h)- 4}+ \sqrt{5x-4}}=\frac{5h}{\sqrt{5(x+h)- 4}+ \sqrt{5x-4}}[/tex]

    Can you finish?
     
  9. Mar 7, 2012 #8

    Mark44

    Staff: Mentor

    But many textbooks present what they call the chain rule form of the power rule.

    ##\frac{d~u^n}{dx} = n\cdot u^{n - 1} \cdot \frac{du}{dx}##

    For example, d/dx(sin3(x)) = 3*sin2(x) * cos(x)
     
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