How would you us the composite rule to differentiate
k(x)=ln(6cos((1/3)x))?
Hello, Orlando!
How would you use the composite rule to differentiate: .$\displaystyle k(x)\:=\:\ln\left(6\cos\tfrac{1}{3}x\right)$
$\displaystyle k'(x) \;=\;\frac{1}{6\cos\tfrac{1}{3}x} \cdot (-6\sin\tfrac{1}{3}x) \cdot\left(\tfrac{1}{3}\right) \;=\;-\tfrac{1}{3}\frac{\sin\tfrac{1}{3}x}{\cos\tfrac{1} {3}x} \;=\;-\tfrac{1}{3}\tan\tfrac{1}{3}x$
$\displaystyle k'(x)=\frac{1}{6 \cdot \cos\left(\frac{1}{3}x\right)} \cdot \left(6 \cdot \cos\left(\frac{1}{3}x\right)\right)'=$
$\displaystyle =\frac{1}{6 \cdot \cos\left(\frac{1}{3}x\right)} \cdot 6 \cdot \left(- \sin \left(\frac{1}{3}x\right)\right) \cdot \frac{1}{3}=\frac{-1}{3} \cdot \tan \left(\frac{1}{3}x\right)$
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))