## clara1223 one year ago find f'(c) of f(x)=sin(6x)/(1+cos(3x)), c=0

1. anonymous

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

2. clara1223

yes

3. anonymous

Ok lets look at the product rule and the chain rule in that order. First I will right it like this (personal preference): $f(x)=sin(6x)*(1+cos(3x))^{-1}$

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

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.

9. anonymous

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

10. anonymous

That shouldn't be too difficult @clara1223

11. clara1223

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

12. clara1223

@PlasmaFuzer ?

13. anonymous

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

14. clara1223

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

15. clara1223

@PlasmaFuzer is that right?

16. anonymous

Yup I agree