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Find the derivative of the functions

  1. Aug 19, 2008 #1
    The question is as below:

    g(x) = (2x2+x+1) / (x2+2x+1)

    It says find the derivative of the functions, I don't get it what does it meant by that, derivative supposed to be dy/dx kind of things right, but all the tutorials is so simple, but how do you do this question??
     
  2. jcsd
  3. Aug 19, 2008 #2

    HallsofIvy

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    Re: derivative

    First, I am going to move this. Just because it is about a derivative, does not mean it has anything to do with differential equations!

    Yes, the derivative is dy/dx or, in this case because the function is called "g", dg/dx.

    Since g(x) is defined as a fraction, do you know the "quotient rule"?
    [tex]\frac{d \frac{f}{g}}{dx}= \frac{\frac{df}{dx}g- f\frac{dg}{dx}}{g^2}[/tex]
     
  4. Aug 19, 2008 #3
    Re: derivative

    No, but i managed to find 2 sites about the question, and i got the final answer in below, i wonder if this is right??

    A: (4x+1) / (2x+2)
     
  5. Aug 19, 2008 #4

    HallsofIvy

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    Re: derivative

    No, it appears that you have just differentiated the numerator and denominator separately and ignored the formula I gave.
    [tex]\frac{d\frac{f}{g}}{dx}\ne \frac{\frac{df}{dx}}{\frac{dg}{dx}}[/tex]

    Your function is g(x) = (2x2+x+1) / (x2+2x+1) which is of the form f(x)/h(x) with f(x)= 2x2+ x+ 1 and h(x)= x2+ 2x+ 1. Then f'(x)= 4x+ 1 and h'(x)= 2x+ 2. The formula I gave
    [tex]\frac{d\frac{f}{h}}{dx}= \frac{\frac{df}{dx}g- f\frac{dg}{dx}}{g^2}[/tex]
    then gives
    [tex]\frac{dg}{dx}= \frac{(4x+1)(x^2+ 2x+1)- (2x^2+ x+ 1)(2x+2)}{(x^2+ 2x+ 1)^2}[/tex]
    [tex]= \frac{3x^2+ 2x-1}{(x^2+ 2x+ 1)^2}[/tex]
    which, I believe does not simplify any more.
     
  6. Aug 19, 2008 #5
    Re: derivative

    If you are not familiar with the quotient rule (which you should change immediately), you might be familiar with the product rule. In which case rewrite

    [tex] g(x) \, = \, \frac{2x^{2} \, + \, x \, + \, 1}{x^{2} \, + \, 2x \, + \, 1} [/tex]

    as

    [tex] g(x) \, = \, \left( 2x^{2} \, + \, x \, + \, 1 \right) \left(x^{2} \, + \, 2x \, + \, 1 \right)^{-1} [/tex]
     
  7. Aug 19, 2008 #6
    Re: derivative

    New question,

    find the derivative of the function,

    f(x) = sin2x(3x2-2)5=cos2x(3x2-2)5 + sin2x(30x-10)4

    Can i simplify until the answer: (3x2-2)5 + (30x-10)4
     
  8. Aug 19, 2008 #7

    HallsofIvy

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    Re: derivative

    I have no idea what you are doing! You are simply handing us purported "answers"- they are in fact not even close to correct- without showing any work.

    Do you know the "chain rule"?
    Do you know what the derivative of sin x is?
     
  9. Aug 19, 2008 #8
    Re: derivative

    The derivative of the sin x is cos x, right??
     
  10. Aug 19, 2008 #9
    Re: derivative

    Yes now do you know the deriviative of [tex] f(g(x)) [/tex]?
     
  11. Aug 19, 2008 #10
    Re: derivative

    how is product rule different from chain rule??
     
  12. Aug 19, 2008 #11
    Re: derivative

    In way that product rule is for

    f(x)*g(x)

    so d/dx [f(x)*g(x)] = f'*g + f*g'
    and chain rule is for

    f[g(x)] = g'(x)*f'[g(x)]
     
  13. Aug 19, 2008 #12
    Re: derivative

    Chain rule: [tex]\frac{d}{dx}[f(g(x))] = g'(x)*f'(g(x))[/tex]

    Product rule: [tex]\frac{d}{dx}[f(x)*g(x)] = f'(x)g(x)+g'(x)f(x)[/tex]

    Can you see the differences? It sounds like your trying to teach yourself calculus starting with derivative shortcuts, that's a bad idea. Try looking in the tutorials section for video lectures/e-books instead of reading tutorials on shortcuts.

    Edit: Damn beat to it XD
     
  14. Aug 19, 2008 #13
    Re: derivative

    I think the first step is understanding the difference betweeen function multiplication and function composition. Do you know that difference from your algebra class?
     
  15. Aug 19, 2008 #14
    Re: derivative

    so the answer should be something like this:

    (30x - 10)4 * cos2(3x2-2)5
     
  16. Aug 19, 2008 #15
    Re: derivative

    >> d/dx '(sin(x))^2*(3*x^2-2)^5'

    ans=

    It's not

    It's 2*sin(x)*(3*x^2-2)^5*cos(x)+30*sin(x)^2*(3*x^2-2)^4*x

    This question seems to be above your level right now, try some simpler questions first for good understanding of diff rules and then try this one ...
     
  17. Aug 19, 2008 #16
    Re: derivative

    isn't it f(x) = sin2x and g(x) = (3x2-2)5?? and how is it possible that it evolve until that stage and how does cos x + 30 came from??
     
  18. Aug 19, 2008 #17
    Re: derivative

    Yes, so
    it looks like
    f[p(x)]*g[q(x)]

    Hint: p(x) = sin(x) and f[p(x)] = p(x)^2
    similarly q(x) = 3x^2-2 and g(x) = q(x)^5

    therefore
    d/dx f[p(x)]*g[q(x)] = ??

    Use chain rule and product rule together
     
  19. Aug 19, 2008 #18
    Re: derivative

    can anyone show me the whole solution to this question, this is my assignment:

    f(x) = sin2x(3x2-2)5

    Please and thank you.
     
  20. Aug 19, 2008 #19
    Re: derivative

    No, I certainly hope no one just shows you the answer, after all it is YOUR assignment. RootX's response should be enough to help you figure it out. What about this derivative is still confusing you? Give it a shot yourself.
     
  21. Aug 19, 2008 #20
    Re: derivative

    Is it:
    f[p(x)]*g[q(x)]

    or

    f'[p(x)]*g'[q(x)]
     
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