Question

find f'(c) of f(x)=sin(6x)/(1+cos(3x)), c=0

Answer 1

So we have:
\[f(x)=\frac{sin(6x)}{1+cos(3x)}\]
correct?

Answer 2

yes

Answer 3

Ok lets look at the product rule and the chain rule in that order.

First I will right it like this (personal preference):
\[f(x)=sin(6x)*(1+cos(3x))^{-1}\]

Answer 4

Apply the product rule:
\[\frac{d}{dx}(f(x))=f'(x)=\frac{d}{dx}(sin(6x)*(1+cos(3x))^{-1}) \\ \ \ \ \ \ = \frac{d}{dx}(sin(6x))*(1+cos(3x))^{-1}+sin(6x)*\frac{d}{dx}((1+cos(3x))^{-1})\]

Answer 5

Now evaluate the derivatives and apply the chain rule on the second term:
\[f'(x)=6cos(6x)*(1+cos(3x))^{-1}+sin(6x)*(-1)*(1+cos(3x))^{-2}*(-3sin(3x))\]

Answer 6

darn....
\[f'(x)=6cos(6x)*(1+cos(3x))^{-1}+ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin(6x)*(-1)*(1+cos(3x))^{-2}*(-3sin(3x))\]

Answer 7

Now lets tidy it up:
\[f'(x)=\frac{6cos(6x)}{1+cos(3x)}-\frac{sin(6x)}{(1+cos(3x))^{2}}*(-3sin(3x)) \\ \ \ \ \ \ \ \ \ \ = 3(\frac{2cos(6x)}{1+cos(3x)}+\frac{sin(6x)sin(3x)}{(1+cos(3x))^{2}})\]

Answer 8

Ok now we can continue to do algebra on this statement and no doubt use trig identities to simplify further, but I am not sure what you require.

Answer 9

Oh wait nm we have a value we need to substitute into it I forgot...

Answer 10

That shouldn't be too difficult @clara1223

Answer 11

If c=0 then wouldnt the answer be 0 because sin(0)=0?

Answer 12

@PlasmaFuzer ?

Answer 13

No only the second term would zero out because the first term contains only cosine terms.... note the multiplier was only for the second term

Answer 14

the first term would be 1, so the final answer is 3?

Answer 15

@PlasmaFuzer is that right?

Answer 16

Yup I agree