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Can anybody differentiate this to simplest formy=4/3+5cosx

  1. Jun 20, 2004 #1
    can anybody differentiate this to simplest form
    y=4/3+5cosx
     
    Last edited: Jun 20, 2004
  2. jcsd
  3. Jun 20, 2004 #2

    arildno

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    What about trying it out yourself?
    What is your specific problem with this?
     
  4. Jun 20, 2004 #3

    HallsofIvy

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    Do you know the derivative of a constant?

    Do you know the derivative of cos(x)?
     
  5. Jun 23, 2004 #4

    nix

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    for 5cos(x) product rule may be useful
     
  6. Jun 23, 2004 #5

    HallsofIvy

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    By the way, is that (4/3)+ 5cos(x) or 4/(3+5cos(x))?
     
  7. Jun 26, 2004 #6

    Kurdt

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    nix: The product rule does not need to be applied as the 5 looks to be independant of x.
     
  8. Jul 19, 2004 #7
    Hello,
    I have a limit to solve which in fact involves differentation (L'Hospital's rule)

    Lim as x->inf. (xe^[1/x]-x)
    I solved it but I am not quite sure. It can't be solved immediately by substitution because it becomes an undet. form. So applying the Hospital Rule I found Lim as x->inf (e^[1/x] - [(e^{1/x})/x] - 1)
    Then it gives e^0 -0-1 = 1-1=0
    I don't know now if it's correct because I don't even have a TI calc. to check the answer.
    Thanx
     
  9. Jul 20, 2004 #8
    Your answering a question that is a month old and that isn't the correct solution either.
    [tex]
    y=\frac{4}{3+5\cos{x}}
    [/tex]

    You can use the quotient rule or change the denomenator to a negative power

    Quotient rule:

    given a function f(x)
    [tex]
    f(x)=\frac{g(x)}{h(x)}
    [/tex]

    Then the quotient rule would be:
    [tex]
    f^\prime (x)=\frac{h(x)g^\prime (x)-h^\prime (x)g(x)}{(h(x))^2}
    [/tex]

    if we let the numerator be g(x) and the denomenator be h(x) we get:
    [tex]
    g(x)=4
    [/tex]

    [tex]
    g^\prime (x)=0
    [/tex]

    [tex]
    h(x)=3+5\cos{x}
    [/tex]

    [tex]
    h^\prime (x)=-5\sin{x}
    [/tex]

    Plugging the individual functions into the quotient rule yields:
    [tex]
    f^\prime (x)=\frac{(3+5\cos{x})(0)-(-5\sin{x})(4)}{(3+5\cos{x})^2}
    [/tex]

    simplify:
    [tex]
    f^\prime (x)=\frac{20\sin{x}}{(3+5\cos{x})^2}
    [/tex]

    If you look at the problem as:
    [tex]
    y=\frac{4}{3}+5\cos{x}
    [/tex]

    you get:
    [tex]
    y^\prime=-5\sin{x}
    [/tex]


    [edit]transposed 3 and 4...fixed
     
    Last edited: Jul 20, 2004
  10. Jul 20, 2004 #9

    HallsofIvy

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    Your answering a question that is a month old and that isn't the correct solution either.
    [tex] y= \frac{4}{3+5cos x} [/tex]
    How did you decide that that was the problem? I would think that y= 4/3+ 5 cos x would be more reasonably interpreted as y= (4/3)+ 5 cos x. I asked lakshmi earlier which was intended and got no answer.
     
  11. Jul 20, 2004 #10
    I answered both possible forms of the question.

    [edit]Also, KnowledgeIsPower you have a misconception about the generel form. the derivative [itex]x^n[/itex] only applies to [itex]x^n[/itex]. You can't apply this to the transcendental functions because each has its own rule. For example: the derivative [itex]\ln x[/itex] is not [itex]1/\ln x[/itex] as [itex]x^n[/itex] would suggest.

    [edit][edit] The response I was addressing seems to have disappeared anyway.
     
    Last edited: Jul 20, 2004
  12. Jul 20, 2004 #11
    Ok thanks for the information. I just studied a number of new trig content and thought i'd jump right in - though i hadn't done differentiation/integration on trigonomic functions.
    My mistake, sorry.

    But what prevents you from just classing sinx as 'y' and differentiating that way?
     
    Last edited: Jul 20, 2004
  13. Jul 20, 2004 #12
    You can say y=sin x but you have to use the chain rule to get dy/dx.

    Here, given: y=sin x and you want to find dy/dx.

    let A=sin x

    y=A
    dy=dA
    dy/dA=1

    Well that's not what we wnat we are trying to find dy/dx

    [tex]
    \frac{dy}{dx}=\frac{dy}{dA}\frac{dA}{dx}
    [/tex]

    The above works just like algebra. The dA's will cancel to give dy/dx; however, you must find dA/dx to use the chain rule.

    A=sin x
    dA=cos xdx
    dA/dx=cos x

    thus dy/dx=(1)(cos x)
    or dy/dx=cos x

    back to the original problem: y=sin x
    You'll learn this soon enough, but the derivative of sin x is cos x so and the derivative of cos x is -sin x.

    thus if we find the derivative of y=sin x we get dy/dx=cos x

    I'd liken the process to fractional addition. You can't simply add 1/3 and 1/2 together. You have to first find a common denominator. Here too we can't simply say sin x=A and get the same result as y=sin x. We still need to do the intermediate steps to convert dA/dx and dy/dA to dy/dx.

    You'll soon learn that there are 20 to 30 common derivatives you'll need to memorize.

    Hope this helped.
     
  14. Jul 21, 2004 #13
    Thanks a lot for the explanation. That was very interesting and informative. Fools rush in..., guess i shouldn't jump ahead.
    It's amazing that you people will help those less knowledgable out for free. It's a pity there's no way to return the favour.
    Thanks again.
     
    Last edited: Jul 21, 2004
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