Chain Rule Help

  • Thread starter dec1ble
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  • #1
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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!
 

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  • #2
saltydog
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dec1ble said:
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!

Break it down:

Suppose have a function of x as u(x) and have the expression:

ln[u(x)]

Would you not, using the chain rule, take the derivative of the logarithm which is [itex]\frac{1}{u(x)}[/itex], then take the derivative of u(x) to get:

[tex]\frac{1}{u(x)}u^{'}[/tex]

Same dif with the cosine of a function right? Derivative is minus the sine of the function times the derivative of the function. In your case, it's nested three times (well w-1 is a function of w in which the derivative is just 1).
 
  • #3
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if i have [tex] \frac{d}{dx} \ln u = \frac{1}{u} [/tex]

then i find d/dx cos(w-1) = -sin(1)

then plug that into the u?

so 1/-sin(1) - then im stuck
 

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