clara1223
  • clara1223
find f'(c) of f(x)=sin(6x)/(1+cos(3x)), c=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
So we have: \[f(x)=\frac{sin(6x)}{1+cos(3x)}\] correct?
clara1223
  • clara1223
yes
anonymous
  • anonymous
Ok lets look at the product rule and the chain rule in that order. First I will right it like this (personal preference): \[f(x)=sin(6x)*(1+cos(3x))^{-1}\]

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anonymous
  • anonymous
Apply the product rule: \[\frac{d}{dx}(f(x))=f'(x)=\frac{d}{dx}(sin(6x)*(1+cos(3x))^{-1}) \\ \ \ \ \ \ = \frac{d}{dx}(sin(6x))*(1+cos(3x))^{-1}+sin(6x)*\frac{d}{dx}((1+cos(3x))^{-1})\]
anonymous
  • anonymous
Now evaluate the derivatives and apply the chain rule on the second term: \[f'(x)=6cos(6x)*(1+cos(3x))^{-1}+sin(6x)*(-1)*(1+cos(3x))^{-2}*(-3sin(3x))\]
anonymous
  • anonymous
darn.... \[f'(x)=6cos(6x)*(1+cos(3x))^{-1}+ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ sin(6x)*(-1)*(1+cos(3x))^{-2}*(-3sin(3x))\]
anonymous
  • anonymous
Now lets tidy it up: \[f'(x)=\frac{6cos(6x)}{1+cos(3x)}-\frac{sin(6x)}{(1+cos(3x))^{2}}*(-3sin(3x)) \\ \ \ \ \ \ \ \ \ \ = 3(\frac{2cos(6x)}{1+cos(3x)}+\frac{sin(6x)sin(3x)}{(1+cos(3x))^{2}})\]
anonymous
  • 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.
anonymous
  • anonymous
Oh wait nm we have a value we need to substitute into it I forgot...
anonymous
  • anonymous
That shouldn't be too difficult @clara1223
clara1223
  • clara1223
If c=0 then wouldnt the answer be 0 because sin(0)=0?
clara1223
  • clara1223
@PlasmaFuzer ?
anonymous
  • 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
clara1223
  • clara1223
the first term would be 1, so the final answer is 3?
clara1223
  • clara1223
@PlasmaFuzer is that right?
anonymous
  • anonymous
Yup I agree

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