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Chain Rule Help

  1. Mar 3, 2005 #1
    I am presenting a problem in front of the class tomorrow and I am slightly confused on the steps for my problem. The problem is:

    Find the derivative of the given function

    f(w) = ln[cos(w-1)]

    The answer in the back of my book shows the derivative is -tan(w-1) - but I'm my steps aren't giving that answer - could anyone show me the steps to use in order to get that answer? Thanks a lot!
     
  2. jcsd
  3. Mar 3, 2005 #2

    dextercioby

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    What's the cain rule for [tex] \frac{d}{dx} \ln u(x) [/tex]...?

    Daniel.
     
  4. Mar 3, 2005 #3
    well the derivative of (ln x) is 1/x
     
  5. Mar 3, 2005 #4

    dextercioby

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    I didn't ask u that...Oh,DO NOT DOUBLE POST...!!!:mad:

    I meant for a logarithm whose argument is a general function u(x),not the particular value u(x)=x...

    Daniel.
     
  6. Mar 3, 2005 #5
    hint: [tex] \frac{d}{dx} \ln u = \frac{1}{u} du [/tex] Now what is u?
     
    Last edited: Mar 3, 2005
  7. Mar 3, 2005 #6
    see thats where im confused.....i believe i have to use the chain rule twice....but im just confused on where to start and how to integrate in the ln
     
  8. Mar 3, 2005 #7
    the u would be the cos(w-1) right ?
     
  9. Mar 3, 2005 #8
    yes. now you have to find the dervative of u and multiply it by your previous result.
     
  10. Mar 3, 2005 #9

    dextercioby

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    Th formula you hinted is incomplete,hence INCORRECT.Please,refrain from erroneous advice...

    Daniel.
     
  11. Mar 3, 2005 #10
    alright - so when i find the derivative of cos(w-1) - it will be -sin(1) - making the equation -sin(1)/cos(w-1), making -sin/cos = -tan(w-1) the answer

    thanks for all your guys help - much appreciated
     
  12. Mar 3, 2005 #11
    To clarify, [tex]\frac{d}{dx}\ln{u} = \frac{1}{u}du[/tex].

    The "du" is very important...

    Daniel... what formula were you referring to as incorrect?
     
  13. Mar 3, 2005 #12

    dextercioby

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    There are 2 now:the first and the last.The correct one is:
    [tex] \frac{d(\ln u)}{dx}=\frac{1}{u}\frac{du}{dx} [/tex]

    Daniel.
     
  14. Mar 3, 2005 #13
    Ah, I see now. Good point.
     
  15. Mar 3, 2005 #14
    The derivative of cos(w-1) is not equal to -sin(1). An easy way for me to understand the chain rule is to say to myself "derivative of the outside times the derivative of the inside". What's the derivative of cos(x)? Now, when you have that, sub in x as w-1, then find the derivative of w-1.
     
  16. Mar 4, 2005 #15

    BobG

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    You're on the right track thinking you need to apply the chain rule twice, but you applied it wrong. The derivative of cos(u) is -sin(u)*du

    In this case, the u=(w-1), and the derivative of (w-1) is 1. You should have -sin(w-1)*1. Now you wind up with:

    [tex]\frac{-sin(w-1)}{cos(w-1)} = -tan(w-1)[/tex]
     
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