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Derivitive problems

  1. Jul 3, 2007 #1
    1. The problem statement, all variables and given/known data
    The directions state, Use the definition to differentiate the functions given.
    1)f(x)=sqrt 5x. the answer states: (square root 5x)/2x.

    The directions state, a. Find the difference quotient of f. and b. f'(c) by comparing the limit of the difference quotient. Also, c is a constant.
    2)f(x)=2-x^2 at c=0. the answer states: for a. -h and b. 0

    2. Relevant equations
    1) sqrt root5x+sqrt root5h-sqrt 5x/h= sqrt5h/h. This is as far as I have gotten.
    2)first, change of x =h, 2-x^2-2xh-h^2-2+x^2/h= h(2x-h)/h=(2x-h).

    3. The attempt at a solution
    For 1), I have not even came close to the answer said in the book.
    For 2), Logically(from other problems i tried), it would be 2x, but as it says above, it is -h and for part b, I believe I have to find part a before I go on to part b. plz help
    Last edited: Jul 3, 2007
  2. jcsd
  3. Jul 3, 2007 #2
    We know that the definition of the derivative for a function f is

    [tex] f^\prime (x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} [/tex]

    so let [tex]f(x) = \sqrt{5x}[/tex]. Now after you've substituted it in, it might look as though you can't do anything with it. We want to get rid of the radicals in the numerator. This can be done by mulitplying by the conjugate of the radical.

    (Edit: Don't forget to multiply both the top and the bottom by the conjugate so that you don't change the equation)

    That is [tex]\displaystyle\left( \sqrt{a}-\sqrt{b} \right) \left( \sqrt{a}+\sqrt{b}\right) = a-b[/tex]. Give this a try and things should work out from there.
    Last edited: Jul 3, 2007
  4. Jul 3, 2007 #3
    i multiplied the rad. on the numerator by (sqrt 5x + sqrt 5h) +(sqrt 5x) (since

    numerator has negative sign), and i got 5x+5h-5x/h(sqrt 5x + sqrt 5h) +(sqrt

    5x). both 5x's on numerator go for it to be 5h/h(sqrt 5x + sqrt 5h) +(sqrt 5x).

    I'm not sure on how to get (sqrt5x)-2x from there, but I know I'm still doing

    something wrong here.
  5. Jul 3, 2007 #4
    Okay, you're close but you seem to have made a bit of an arithmetic error. You should've had

    [tex]\frac{\sqrt{5x+5h}-\sqrt{5x}}{h} \frac{\sqrt{5x+5h}+\sqrt{5x}}{\sqrt{5x+5h}+\sqrt{5x}}[/tex]

    This then simplifies to

    [tex]\displaystyle\frac{5h}{h \left(\sqrt{5x+5h}+\sqrt{5x}\right )}[/tex]

    Edit: Don't forget that the h is multiplied by the entire conjugate, and that addition doesn't distribute over a radical. Thus [tex]\sqrt{5(x+h)} \neq \sqrt{5x}+\sqrt{5h}[/tex] which is what you have written down.
    Last edited: Jul 3, 2007
  6. Jul 3, 2007 #5
    actually that is what I was trying to say, didn't know how to write it. now I am aware the h cancels, for it to be, 5/(sqrt 5x+5h +sqrt5x). I know I cannot add (sqrt 5x+5h +sqrt5x) together because it is not like. If I am wrong about that, let me know. Am I suppose to distribute it back? I don't know about any other options than this step.
  7. Jul 3, 2007 #6
    Okay, so assuming that you have what I had in my previous post, take the limit as [tex]h\to 0[/tex]. You won't get the stated answer, but the two are equivalent. What you should get is


    But [tex]\frac{5}{\sqrt{5}}=\sqrt{5}[/tex] giving [tex]\frac{\sqrt{5}}{2\sqrt{x}}[/tex].

    Now mulitply top and bottom by [tex]\sqrt{x}[/tex] to get [tex]\frac{\sqrt{5x}}{2x}[/tex] as required.
  8. Jul 3, 2007 #7
    It may seem like we went through a lot of steps, but this is a method that can be applied anytime you have a function with a radical in it. This is also a very popular exam question since using the definition of the derivative on anything larger than a quadratic polynomial is tedious, and many of the other elementary functions require identities that you may not have learned or that are not taught in the course. Just remember, when dealing with radicals:

    Write it out using the definition
    Multiply by the conjugate
    Let [tex]h \to 0[/tex]

    Following those steps guarantees you'll get the problem everytime.
  9. Jul 3, 2007 #8
    Thank you, I will keep that in mind.
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