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Derivatives, and a little bit of linear approxim

  1. Jul 4, 2014 #1
    1. The problem statement, all variables and given/known data
    I have to find the deriv
    of ##f(x)=xArctan(x^{3})##

    I just need an explanation of how the arctan works...
    So I understand the rest, but just the deriv of arctan itself is confusing for me.
    So the derivative of ##arctan(x)## just in general is ##\frac{1}{1+x^{2}}##

    But if it's of ##arctan(x^{3})##
    Then I'm not sure what is right

    So on my test I was pretty sure that when I derived arctan, that the x^2 factor meant (x)^2, so I would have to put in ##x^3##, and then I'd get ##\frac{1}{1+x^{6}}##
    (That's x to the 6th, because you'd mult the exps)

    But.... my test says I forgot to chain rule the ##x^{3}## that's inside of arctan(), but why did the teacher write (.3x^2) as the ##g'(x)##?


    But I also wrote ##\frac{1}{1+3x^{4}}##, which wasn't marked wrong as the deriv of ##arctan(x^{3})##, so.. I'm confused.
     
  2. jcsd
  3. Jul 4, 2014 #2

    SteamKing

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    If y = tan (x), then x = arctan (y), that is, arctan is the inverse function of tan.

    For d/dx [arctan (x^3)], you should make the substitution u = x^3,

    then d/dx [arctan (x^3)] = d/du [arctan (u)] * du/dx, by the chain rule.

    The derivative of arctan u = 1/(1+u^2) [from the formula] and I'm sure you can calculate du/dx, when u = x^3
     
  4. Jul 4, 2014 #3
    So.. wait I'm sorry, so then what's the formula?
    On your last line I'm not so clear what you mean, sorry.
     
  5. Jul 4, 2014 #4

    SteamKing

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    The last line just repeats the formula for the derivative of the arctan, with the argument of arctan being 'u' instead of 'x'. Then, because you differentiated arctan (u) with respect to 'x', and you know the relationship between 'u' and 'x', u = x^3, the chain rule calls for you to calculate du/dx and multiply that result by the derivative of arctan (u) to obtain the derivative of arctan (x^3).

    I think you need to review the chain rule. It's going to come in handy if you continue to study calculus and take up integrals.
     
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