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clara1223

  • one year ago

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

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  1. anonymous
    • one year ago
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    So we have: \[f(x)=\frac{sin(6x)}{1+cos(3x)}\] correct?

  2. clara1223
    • one year ago
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    yes

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

  7. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Oh wait nm we have a value we need to substitute into it I forgot...

  10. anonymous
    • one year ago
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    That shouldn't be too difficult @clara1223

  11. clara1223
    • one year ago
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    If c=0 then wouldnt the answer be 0 because sin(0)=0?

  12. clara1223
    • one year ago
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    @PlasmaFuzer ?

  13. anonymous
    • one year ago
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    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
    • one year ago
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    the first term would be 1, so the final answer is 3?

  15. clara1223
    • one year ago
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    @PlasmaFuzer is that right?

  16. anonymous
    • one year ago
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    Yup I agree

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