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clara1223

  • one year ago

find f'(x) of f(x)=sin(2x)(sqrt(1+cos(5x)))

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    product plus chain rule \[(fg)'=f'g+g'f\] with \[f(x)=\sin(x), f'(x)=\cos(x), g(x)=\sqrt{1+\cos(5x)}\] \[g'(x)=\frac{-5\sin(5x)}{2\sqrt{1+\cos(5x)}}\]

  3. clara1223
    • one year ago
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    what happened to the 2x inside sin(x)?

  4. misty1212
    • one year ago
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    oops \[f(x)=\sin(2x), f'(x)=2\cos(2x)\] my bad

  5. clara1223
    • one year ago
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    How would I apply the chain rule to that? The sqrt makes it pretty messy.

  6. clara1223
    • one year ago
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    @misty1212 ?

  7. misty1212
    • one year ago
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    \[\left(\sqrt{f}\right)'=\frac{f'}{2\sqrt{f}}\]

  8. misty1212
    • one year ago
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    which should explain \[g'(x)=\frac{-5\sin(5x)}{2\sqrt{1+\cos(5x)}}\]

  9. clara1223
    • one year ago
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    That makes sense. I'm just not sure what to do from there. From the product rule I got 2cos(2x)(sqrt(1+cos(5x)))+sin(2x)((-5sin(5x))/2(sqrt(1+cos(5x))))

  10. misty1212
    • one year ago
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    leave it

  11. clara1223
    • one year ago
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    that's my final answer?

  12. misty1212
    • one year ago
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    it is a silly made up question anyway what else can you do?

  13. misty1212
    • one year ago
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    you sure as hell don't want to actually add them although you could if you had like an extra half hour to waste

  14. clara1223
    • one year ago
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    Haha definitely not. Still got 6 more questions on this homework sheet due tomorrow.

  15. misty1212
    • one year ago
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    best get busy

  16. misty1212
    • one year ago
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    lol not that "get busy"

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