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Second derivative of xtanx?

  1. Mar 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Given y=xtanx, find y'' (second derivative)

    2. Relevant equations

    Uh... I'm not even sure if I'm using the right one....
    d/dx(tanx) = sec^2x

    3. The attempt at a solution

    y'= (x)(sec^2(x)) + (tanx)(1)
    y'= xsec^2(x) + tanx

    y'' = [(x)(2sec^3(x)) + sec^2(x)(1)] + sec^2x
    y'' = 2xsec^3(x) + sec^2(x) + sec^2(x)


    The key answer says it is supposed to be (2cosx + 2xsinx)/(cos^3(x))
  2. jcsd
  3. Mar 7, 2008 #2
    you took the 2nd derivative wrong

    [tex]y'=x\sec^{2}x+\tan x[/tex]

    [tex]y''=x\cdot2\sec^{2-1}x\cdot\sec x\tan x+\sec^{2}x+\sec^{2}x[/tex]
    [tex]=2x\sec^{2}x\tan x+2\sec^{2}x[/tex]

    What you did was increase the power rather than decreasing it.

    In general, the derivative of secant raised to a power is ... [tex]\frac{d}{dx}(\sec^{n}x)=n\sec^{n}x\tan x[/tex]
    Last edited: Mar 7, 2008
  4. Mar 7, 2008 #3
    Oh wow, I'm an idiot..

    So is the general rule d/dx(sec^n(x)) = nsec^n(x)tan(x) just a combination of the (x^n)' = nx^x-1 and d/dx(secx) = secxtanx?

    What if I were to be givin d/dx(tan^n(x))... would the answer be like nsec^2(n-1)(x)? nsec^n(x)? ...

    Thanks so much for your help, by the way :).
  5. Mar 7, 2008 #4
    No, don't forget the chain rule!!!

  6. Mar 7, 2008 #5
    The chain rule?
    ... I've never actually seen that before.
    Haha, I guess that explains a few things! I haven't been taught that yet.

    Guess I'll go read up on that, and thanks again for your help! I really appreciate it :).
  7. Oct 2, 2011 #6
    i got the + sec^2x but how did you get the other + sec^2x
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