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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we have: \[f(x)=\frac{sin(6x)}{1+cos(3x)}\] correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok lets look at the product rule and the chain rule in that order. First I will right it like this (personal preference): \[f(x)=sin(6x)*(1+cos(3x))^{1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Apply the product rule: \[\frac{d}{dx}(f(x))=f'(x)=\frac{d}{dx}(sin(6x)*(1+cos(3x))^{1}) \\ \ \ \ \ \ = \frac{d}{dx}(sin(6x))*(1+cos(3x))^{1}+sin(6x)*\frac{d}{dx}((1+cos(3x))^{1})\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now evaluate the derivatives and apply the chain rule on the second term: \[f'(x)=6cos(6x)*(1+cos(3x))^{1}+sin(6x)*(1)*(1+cos(3x))^{2}*(3sin(3x))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0darn.... \[f'(x)=6cos(6x)*(1+cos(3x))^{1}+ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin(6x)*(1)*(1+cos(3x))^{2}*(3sin(3x))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now lets tidy it up: \[f'(x)=\frac{6cos(6x)}{1+cos(3x)}\frac{sin(6x)}{(1+cos(3x))^{2}}*(3sin(3x)) \\ \ \ \ \ \ \ \ \ \ = 3(\frac{2cos(6x)}{1+cos(3x)}+\frac{sin(6x)sin(3x)}{(1+cos(3x))^{2}})\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh wait nm we have a value we need to substitute into it I forgot...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That shouldn't be too difficult @clara1223

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If c=0 then wouldnt the answer be 0 because sin(0)=0?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No only the second term would zero out because the first term contains only cosine terms.... note the multiplier was only for the second term

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the first term would be 1, so the final answer is 3?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@PlasmaFuzer is that right?
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