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One regular trig question and one question on derivative of trig function

  1. Jul 3, 2011 #1
    Hey guys these aren't math exercises; I just don't understand a couple parts in my textbook.

    1. Cos(x)/Sin(x) = Cot(x), but Cot(x) * sin(x) ≠ cos(x). Why?

    I know tan(x) * sin(x) ≠ cos(x) because during precalculus nobody ever used sin(x) * cot(x) = cos(x) for anything, but I don't know why you can't do this operation. I think it might have something to do with a 0 in a denominator?

    2. This is a harder question I think. In one section of my book the derivatives of trig functions are defined like this:

    d/dx [sinx] = cos(x)
    d/dx [tanx] = secˆ2(x)
    etc....

    BUT, a couple sections later in the textbook when the chain rule is introduced the definitions of the derivatives of the trig functions change:

    d/dx [sinx] = cos(x)* x'
    d/dx [tanx] = secˆ2(x) * x'
    d/dx [secx] = sec(x)tan(x) * x'
    etc....

    Explain plz :)

    Thanks
     
    Last edited: Jul 3, 2011
  2. jcsd
  3. Jul 3, 2011 #2

    SammyS

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    1.
    Well, for one thing Cos(x)/Sin(x) = Cot(x) , not Tan(x).

    So, Cot(x) * Sin(x) = Cos(x) .


    2.

    x' = d/dx (x) = 1.
     
  4. Jul 3, 2011 #3
    The first part is just about identities,
    You have cos/sin = tan, I think you mean sin/cos = tan. They are basic identities.

    The second part, the reason they have the x' at the end is due to the chain rule. The trig functions are composite functions. If you take the derivative with respect to the angle in the functions you still multiply the derivative of the inside in, it's just that the derivative of the angle with respect to the angle is 1!

    For instance
    [tex]\frac{d}{dx}sin(x) = cos(x)(1)[/tex]
    and
    [tex]\frac{d}{dt}sin(x) = cos(x)x'[/tex]
     
  5. Jul 3, 2011 #4
    Thanks for the replies guys, (sorry about the mistake; I just corrected it) but I'm still confused about both questions.

    1. I realize I made the mistake but I still don't understand why I can't multiply the denominator by the cot(x) on the right side of the equation to get cos(x).

    2. Are you guys saying that the "x" in the second example can represent 10z, so d/dz [sin10z] = 10cos10z, while the x in the first example can't represent anything besides x?

    Sorry again for the mistake.
     
  6. Jul 3, 2011 #5
    No, because it's the derivative with respect to z, it would just be cos(10z).

    Suppose I have some function
    [tex]f(x) = (x^{2}+2)^3[/tex]
    If you take the derivative with respect to x, you get:
    [tex]f'(x) = 3(x^{2}+2)^2(\frac{d}{dx}(x^{2}+2))[/tex]
    Think of (inside)^3 as being one function, and "inside" as being another, then you can see that it is a composition.

    So you take the derivative of the first function:
    3(inside)^2
    and then multiply it by the derivative of what was on the inside to get:
    3(inside)^2 (derivative of inside)

    It's the same thing with the trig functions. It's a composition with some other stuff inside it. So you take the derivative of the outside part, and then multiply the whole thing by the derivative of what is on the inside.

    Here are some examples of the chain rule being used for polynomials:
    http://www.khanacademy.org/video/the-chain-rule?playlist=Calculus [Broken]

    The same thing applies to trig functions.
     
    Last edited by a moderator: May 5, 2017
  7. Jul 3, 2011 #6
    I just typed in to wolframalpha: "derivative of sin(10z)"and it told me the answer is 10cos10z
     
  8. Jul 3, 2011 #7

    SammyS

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    Yes! You were correct!

    d/dz (sin(10z)) = 10 cos(10z), & you're right about this being the point about the chain rule form for the derivatives.
     
  9. Jul 3, 2011 #8
    The guy above you said it is equal to cos10z
     
  10. Jul 3, 2011 #9

    SammyS

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    The only way that cos(x) is different from cot(x) * sin(x), is that the latter is not defined when cot(x) is not defined, that is when x is an integer multiple of π.
     
  11. Jul 3, 2011 #10

    SammyS

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    That's why I replied to you as soon as I saw his post.

    I've made a few mistakes on this site myself. D'oh !
     
  12. Jul 3, 2011 #11
    OK. So far I think I have a little more understanding. Basically, Sammy you're a legend because you have explained the identity thing and I now know why we weren't allowed to substitute cos(x) * tan(x) for sin(x) in trig proofs.

    I am still confused about my second question. My book used the letter "x" for the first derivative I mentioned with the trig functions and a "u" for the second definition. Does the "u" signify any function or number while the "x" signifies a single, untouched, variable? This seems to be the only logical explanation I can think of. Even if this is the case, the book seemed to not explain this...
     
  13. Jul 3, 2011 #12

    SammyS

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    Thanks!

    For your second question, that's about right. The letter "x" is used so often for the universal "independent variable", so the more general derivative rules are written with the variable "u" and thus include the du/dx factor.
     
  14. Jul 3, 2011 #13
    Sammy = Legend + 2

    Ok, so one last question. In precalc we NEVER were allowed to use cos(x)*tan(x) = sin(x) in our proofs. (Now I understand why.) Nevertheless, theoretically, if we wanted to, could I have used cos(x) * tan(x) as a substitute for sin(x) in a proof as long as I specified the domain of the equation at the end of all my operations?
     
  15. Jul 3, 2011 #14

    SammyS

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    Yes.
     
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