# Thread: Using the composite rule to differentiate?

1. ## Using the composite rule to differentiate?

How would you us the composite rule to differentiate

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

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

Hello, Orlando!

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

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

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

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

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

$=\frac{1}{6 \cdot \cos\left(\frac{1}{3}x\right)} \cdot 6 \cdot \left(- \sin \left(\frac{1}{3}x\right)\right) \cdot \frac{1}{3}=\frac{-1}{3} \cdot \tan \left(\frac{1}{3}x\right)$

4. ## 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))