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Homework Help: Normal integral formula problem

  1. Sep 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Given that f'(x)=(cosx)^ (1/3) then find out dy/dx f(ln cosx)=?

    The answer given is -(cos(lncos)) ^ (1/3) tanx

    Anyone can solve it? I tried and got lost with it


    2. Relevant equations

    Using the normal integral formula and derieve it again


    3. The attempt at a solution
     
    Last edited: Sep 16, 2010
  2. jcsd
  3. Sep 16, 2010 #2

    hunt_mat

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    Re: Calculus

    By the chain rule we have:
    [tex]
    f(x)=\cos^{\frac{1}{3}}x=u^{\frac{1}{3}}
    [/tex]
    Use the chain rule to obtain
    [tex]
    \frac{df}{dx}=\frac{df}{du}\frac{du}{dx}
    [/tex]
     
  4. Sep 16, 2010 #3

    HallsofIvy

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    Re: Calculus

    It doesn't really take a genius, just perseverence and steady, step by step work.

    By the chain rule, the derivative of f(ln(cos(x))) is f'(ln(cox(x))) times the derivative of ln(cos(x)). And that is the 1/cos(x) times the derivative of cos(x) which is -sin(x).

    The derivative of f(ln(cos(x))) is f'(ln(cos(x))) times -sin(x)/cos(x).

    Since [itex]f'(x)= (cos(x))^{1/3}[/itex], [/itex]f'(ln(cos(x))= (cos(ln(cos(x)))))^{1/3}[/itex].

    Putting those together gives exactly what your book says.
     
  5. Sep 16, 2010 #4
    Re: Calculus

    here are my further works

    I found out my problem it cant be done

    The problem is i cant increase the power of cos x to (cos x)^2 so there is only one shortest way to solve it
     

    Attached Files:

    Last edited: Sep 17, 2010
  6. Sep 16, 2010 #5
    Re: Calculus

    Here is the shortest way

    f'(x )=cos(x )^(1/3)
    f'(ln(cos(x )))=cos(ln(cos(x )))^1/3* d(ln(cos(x )))/dy* d(cos(x ))/dy
    =cos(ln(cos(x )))^1/3 * 1/cos( x) * (-sin(x ))
    =cos(ln(cos(x )))^1/3 * -sin(x )/cos(x )
    =cos(ln(cos(x )))^1/3 * -tan(x )

    here is the shortest way to finish it

    done by anexgenesis
     
    Last edited: Sep 17, 2010
  7. Sep 16, 2010 #6

    hunt_mat

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    Re: Calculus

    If
    [tex]u=\cos x[/tex]
    and
    [tex]f(u)=u^{\frac{1}{3}}[/tex]
    then
    [tex]df/du=u^{-\frac{2}{3}}/3[/tex]
    that is your problem.
     
  8. Sep 16, 2010 #7

    vela

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    Re: Calculus

    You're misreading the problem. He's given df/dx=(cos x)1/3, not f(x).
     
  9. Sep 16, 2010 #8

    hunt_mat

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    Re: Calculus

    Oh right. I didn't see the dash!
     
  10. Sep 17, 2010 #9
    Re: Calculus

    Problem solved :)
     
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