How would you us the composite rule to differentiate

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

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- Mar 9th 2012, 05:20 AMOrlandoUsing the composite rule to differentiate?
How would you us the composite rule to differentiate

k(x)=ln(6cos((1/3)x))? - Mar 9th 2012, 05:38 AMSorobanRe: Using the composite rule to differentiate?
Hello, Orlando!

Quote:

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

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

- Mar 9th 2012, 05:45 AMprincepsRe: Using the composite rule to differentiate?
$\displaystyle k'(x)=\frac{1}{6 \cdot \cos\left(\frac{1}{3}x\right)} \cdot \left(6 \cdot \cos\left(\frac{1}{3}x\right)\right)'=$

$\displaystyle =\frac{1}{6 \cdot \cos\left(\frac{1}{3}x\right)} \cdot 6 \cdot \left(- \sin \left(\frac{1}{3}x\right)\right) \cdot \frac{1}{3}=\frac{-1}{3} \cdot \tan \left(\frac{1}{3}x\right)$ - Mar 9th 2012, 06:04 AMOrlandoRe: 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))