1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Science Advisor

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    User Avatar
    Homework Helper

    Re: Calculus

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

    Problem solved :)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook