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Derivative using chain rule

  1. Jul 5, 2008 #1
    1. The problem statement, all variables and given/known data
    y= squareroot tan(sin^2 x)


    2. Relevant equations
    chain rule



    3. The attempt at a solution
    f(x)= sqaureroot tan x
    g(x)= (sinx)^2
    f'(x)=1/2 sec^2x ^1/2
    g'(x)= 2 * sinx * cosx

    I dont know if my f'(x) is right if it is then do i just do the chain rule?
     
  2. jcsd
  3. Jul 5, 2008 #2
    [tex]F(x)=\sqrt{\tan{\sin^2 x}}[/tex]

    [tex]f(x)=\sqrt x[/tex]
    [tex]g(x)=\tan x[/tex]
    [tex]h(x)=\sin^2 x[/tex]

    [tex]F(x)=f\{g[h(x)]\}[/tex]
     
  4. Jul 5, 2008 #3

    Hootenanny

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    I'm afraid your f'(x) isn't correct, your g'(x) however is. There is no harm in using the chain rule one more time to make life a little easier. For example, let:

    [tex]g(x) = \sin^2x[/tex]

    [tex]f(g) = \tan\left(g\right)[/tex]

    [tex]h(f) = \sqrt{f}[/tex]

    Then,

    [tex]\frac{d}{dx}\sqrt{\tan\left(\sin^2x\right)} = \frac{dh}{df}\frac{df}{dg}\frac{dg}{dx}[/tex]

    Do you follow?

    Edit: It seems I was beaten to it.
     
    Last edited: Jul 5, 2008
  5. Jul 5, 2008 #4
    yeah kinda

    1/2x(-1/2)tan(sin^2x)[sec^2x(sin^2x)](2sinxcosx)



    is that makin any progress?
     
  6. Jul 5, 2008 #5

    Hootenanny

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    You're not far off. Are you sure that x should be an x?
     
  7. Jul 5, 2008 #6
    is it just 1/2tan(sin^2x)

    i just thought because f'(x)= 1/2x-1/2
     
  8. Jul 5, 2008 #7

    Hootenanny

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    Correct, instead of:

    [tex]\frac{1}{2}x^{-1/2}\ldots[/tex]

    it should be

    [tex]\frac{1}{2}\tan^{-1/2}\left(\sin^2x\right)\ldots[/tex]
    No it isn't.
     
    Last edited: Jul 5, 2008
  9. Jul 5, 2008 #8
    1/2tan^1/2(sin^2x)[sec^2x(sin^2x)](2sinxcosx)
    thats it?
     
  10. Jul 6, 2008 #9

    Hootenanny

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    Careful with your exponent of the tangent, and don't forget that you can cancel the 1/2 with the 2.
     
  11. Jul 6, 2008 #10
    tan^-1/2(sinx)...
     
  12. Jul 6, 2008 #11
    or you mean the 2 in 2sinxcosx
     
  13. Jul 6, 2008 #12
    tan^-1/2(sin^2x)[sec^2x(sin^2x)](sinxcosx)
     
  14. Jul 6, 2008 #13

    Hootenanny

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    Spot on :approve:
     
  15. Jul 6, 2008 #14
    thanks for the help dude
     
  16. Jul 6, 2008 #15

    Hootenanny

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    No problem :smile:
     
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