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The derivative of sin 5x

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data

    what is the derivative of sin 5x



    3. The attempt at a solution

    using the product rule, i think the answer would be

    cos 5x + sin 5

    but the book says the answer is 5 cos 5x
     
  2. jcsd
  3. Jan 29, 2012 #2
    Take the derivative of sin(x) = cos(x), then take derivative (using power rule) of the inside.
    So, derivative of 5x = 5.
    Thus, derivative of sin(anything) = cos(anything) * derivative of anything.
     
  4. Jan 29, 2012 #3
    Differenciating Sin and Cos is very easy. (It gets trickier with the more complicated trig functions, like Tan). The way I like to remember it, is as follows:

    Sin(x)
    Cos(x)
    -Sin(x)
    -Cos(x)

    When you differenciate one of these functions, you simply move down the tower to the next function. When you differenciate -Cos(x), you go back up to Sin(x). Except in this case, as you have (5x) inside the brackets, not just (x). So as you're differenciating, you change the Sin(5x) to Cos(5x) and then multiply by the differential of what's inside the brackets. Here, 5x is inside the brackets, the differencial of which is 5. So you multiply by 5. Hence; 5 Cos(5x).
     
  5. Jan 29, 2012 #4
    [itex]\frac{d}{dx}sin(5x)[/itex]

    Chain Rule:
    Derivative of inner function times the derivative of the outer function.

    [itex]f(g(x)) = f'(g(x))g'(x)[/itex]

    Since
    [itex]g'(x) = \frac{d}{dx}5x = 5[/itex]

    And
    [itex]f'(x) = cos(5x)[/itex]

    Thus:
    [itex]5cos(5x)[/itex]
     
    Last edited: Jan 29, 2012
  6. Jan 29, 2012 #5

    SammyS

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    You are looking at this as the product of what two quantities ?
     
  7. Jan 29, 2012 #6
    I see the mistake I was making. As Sammy pointed out I wrongly thought that cos 5x is two quantities when it's not.
     
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