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Math Help - Using the composite rule to differentiate?

  1. #1
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    shropshire
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    Using the composite rule to differentiate?

    How would you us the composite rule to differentiate

    k(x)=ln(6cos((1/3)x))?
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  2. #2
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    Re: Using the composite rule to differentiate?

    Hello, Orlando!

    How would you use the composite rule to differentiate: . k(x)\:=\:\ln\left(6\cos\tfrac{1}{3}x\right)

    k'(x) \;=\;\frac{1}{6\cos\tfrac{1}{3}x} \cdot (-6\sin\tfrac{1}{3}x) \cdot\left(\tfrac{1}{3}\right) \;=\;-\tfrac{1}{3}\frac{\sin\tfrac{1}{3}x}{\cos\tfrac{1}  {3}x} \;=\;-\tfrac{1}{3}\tan\tfrac{1}{3}x

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  3. #3
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    Re: Using the composite rule to differentiate?

    Quote Originally Posted by Orlando View Post
    How would you us the composite rule to differentiate

    k(x)=ln(6cos((1/3)x))?
     k'(x)=\frac{1}{6 \cdot \cos\left(\frac{1}{3}x\right)} \cdot \left(6 \cdot \cos\left(\frac{1}{3}x\right)\right)'=

    =\frac{1}{6 \cdot \cos\left(\frac{1}{3}x\right)} \cdot 6 \cdot \left(- \sin \left(\frac{1}{3}x\right)\right) \cdot \frac{1}{3}=\frac{-1}{3} \cdot \tan \left(\frac{1}{3}x\right)
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  4. #4
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    Re: Using the composite rule to differentiate?

    thanks for the quick reply.

    The way I have done it has given a completly wrong answer. would you mind showing us where I went wrong

    This is the meathod I used to get to my answer

    k(x)=ln(6cos((1/3)x))

    so k(x)=g(f(x)) where u = f(x) = 6cos((1/3)x) and g(u) = ln u

    so f '(x)= -2 sin (x/3) and g ' (u)= ln u

    so k' (x)= g'(f(x))f '(x)

    k '(x)= g '(u)f '(x) where u = f(x)

    k'(x)= (1/u)(-2sin (x/3))

    k '(x)= (-2sin(x/3))/(6cos((1/3)x))
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