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anonymous

  • one year ago

Find the coefficient of the squared term in the simplified form for the second derivative., f"(x) for f(x)=(X^3+2x+3)(3x^3-6x^2-8x+1). use the hyphen symbol,-, for negative values.

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  1. zepdrix
    • one year ago
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    Hey Maggy :)\[\large\rm f(x)=(x^3+2x+3)(3x^3-6x^2-8x+1)\]Hmm this one looks like a doozy. Remember your product rule?

  2. zepdrix
    • one year ago
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    \[\rm \color{royalblue}{f'(x)}=\color{royalblue}{(x^3+2x+3)'}(3x^3-6x^2-8x+1)+(x^3+2x+3)\color{royalblue}{(3x^3-6x^2-8x+1)'}\]This is how we would "set up" our first derivative, ya? :d

  3. zepdrix
    • one year ago
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    Alternatively, you could expand out all the brackets from the start. That might actually be easier :) Hmm I'm not sure.

  4. zepdrix
    • one year ago
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    ya ya ya ya, expand first. that seems way better >.< Otherwise we have to combine a whole lot of term at the end.. my bad

  5. zepdrix
    • one year ago
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    we allowed to do that, ya? :d

  6. anonymous
    • one year ago
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    my head hurts trying to follow you lol so we expand?

  7. zepdrix
    • one year ago
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    ya let's try that instead :) product rule will result in two groups of stuff. second derivative will require product rule twice more. then expanding three sets of things, and combining. expanding first seems way simpler

  8. zepdrix
    • one year ago
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    |dw:1438993951176:dw|

  9. zepdrix
    • one year ago
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    Here is the expansion of the first two terms from the first set of brackets multiplying everything in the other brackets. Does it kinda make sense what I'm doing? :d there is one more term to take care of.

  10. anonymous
    • one year ago
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    yes I see you expanding to dind the derivatives. o.o math is not my forte lol

  11. zepdrix
    • one year ago
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    |dw:1438994231877:dw|we have another set of terms to get. so we'll have to distribute this 3.

  12. zepdrix
    • one year ago
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    |dw:1438994268041:dw|

  13. zepdrix
    • one year ago
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    We're going to take two derivatives. If we want to END UP WITH a second power. which power of x do you think we'll be starting with? :)

  14. anonymous
    • one year ago
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    with 9x^3?

  15. zepdrix
    • one year ago
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    when we take the derivative of a third power such as that term, we'll get a second power. when we take the derivative again, it degrades down to first power, ya?

  16. zepdrix
    • one year ago
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    So it looks like we'll want to pay attention to the `fourth power` term. That is the one which becomes `second power` after differentiating it twice.

  17. zepdrix
    • one year ago
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    |dw:1438994442968:dw|

  18. zepdrix
    • one year ago
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    You `could` take the derivative of the entire thing. But it doesn't look the problem is requiring that, if I'm reading it correctly. They simply want the coefficient on the 2nd power term in your second derivative. So we currently have -2x^4. What do you get when you differentiate -2x^4 twice?

  19. anonymous
    • one year ago
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    -2x^8??? ( when you say twice you mean add or multiply im getting very confused)

  20. zepdrix
    • one year ago
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    \(\large\rm (-2x^4)'=?\)

  21. zepdrix
    • one year ago
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    Remember your power rule? :)

  22. anonymous
    • one year ago
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    sorta but not confidently

  23. zepdrix
    • one year ago
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    \(\large\rm (x^n)'=n x^{n-1}\) The exponent comes down in front to multiply, and the exponent decreases by 1.

  24. zepdrix
    • one year ago
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    Example: \(\large\rm (4x^7)'=4\cdot7x^{7-1}=28x^{6}\) See how the 7 comes down in front, and the power changes to a 6?

  25. anonymous
    • one year ago
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    so it be 2x^3? from 2x^4?

  26. zepdrix
    • one year ago
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    Nooo :O Where did your 4 go? it didn't come down silly!

  27. anonymous
    • one year ago
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    I cant see your last posted picture that wy I ask my answer was right lol all I see in your last picture posst is symbols

  28. zepdrix
    • one year ago
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    oh the math code didn't format correctly :( ugh

  29. zepdrix
    • one year ago
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    This was the example:|dw:1438995007902:dw|

  30. zepdrix
    • one year ago
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    The 7 comes down, and then the power decreases to a 6.

  31. anonymous
    • one year ago
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    -8x^3? my reasoning: -2x^4= -2. 4x^3 = -8x3

  32. zepdrix
    • one year ago
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    Oo ok nice nice nice. That takes care of our first derivative. Now we have to repeat the process! :)|dw:1438995233431:dw|

  33. anonymous
    • one year ago
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    -8x^3 -8x^3=-8.3x^2=-24x^2?

  34. Jhannybean
    • one year ago
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    \[\checkmark\]

  35. zepdrix
    • one year ago
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    Yayyy good job \c:/ Notice that the exponent we started with was a 4. And the one we ended with is a 2. Differentiating twice made the power decrease by 2. That will happen to every other term as well. So our 5th power term will becomes a 3rd power, our 3rd power term will become a 1st power, and so on. So we only needed to differentiate the 4th power term twice, because that's the only one giving us 2nd power!

  36. zepdrix
    • one year ago
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    So yay good, you've found your coefficient! :) -24

  37. anonymous
    • one year ago
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    ok so my answer to my question is -24x^2?

  38. zepdrix
    • one year ago
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    The question said `Find the coefficient`. The coefficient is `only` the number in front.

  39. anonymous
    • one year ago
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    oh coolz hehe can you help me in one more question?

  40. zepdrix
    • one year ago
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    bahh i need a maf break :) open up a new thread so people can find your question faster! this one is a mess now hehe

  41. anonymous
    • one year ago
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    lol ok ty

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