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anonymous

  • 4 years ago

Think my textbook may have the wrong answer for this: Find all higher derivatives -> f(r) = r(4r + 9)^3 I've tried a few different things, none are giving me the textbook answer. Please show explanation and not just the answer.

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  1. anonymous
    • 4 years ago
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    I see you typing dumbcow. You're awesome :)

  2. dumbcow
    • 4 years ago
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    this is combination of product rule and chain rule let u = r, v = (4r+9)^3 u' = 1 v' = 4*3(4r+9)^2 = 12(4r+9)^2 f'(r) = u'v + uv' f'(r) = (4r+9)^3 + 12r(4r+9)^2 ------------------------ for 2nd derivative let g(r) = 12r(4r+9)^2, we'll find g' first u = 12r, v = (4r+9)^2 u' = 12 v' = 4*2(4r+9) = 8(4r+9) g'(r) = u'v + uv' g'(r) = 12(4r+9)^2 + 96r(4r+9) --> f''(r) = 4*3(4r+9)^2 + 12(4r+9)^2 + 96r(4r+9) f''(r) = 24(4r+9)^2 + 384r^2 + 864r --------------------------- for 3rd derivative, f'''(r) = 48(4r+9) + 768r + 864 ----------------------------------- for 4th derivative f''''(r) = 192 + 768 = 960

  3. dumbcow
    • 4 years ago
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    depending on how the textbook simplifies answers, it might be same thing but look different is everything expanded out in their answers?

  4. anonymous
    • 4 years ago
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    Here's the first two: \[f \prime(r) = (16r+9)(4r+9)^{2}, f ^{n}(r) = 24(8r+9)(4r+9)\] It's strange because I went ahead and did the question after it and had no problems, although it is a slightly different question (4 sqrt(8x-3))

  5. anonymous
    • 4 years ago
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    Yeah I'm getting the same answer as you are, I'll leave a note asking my teacher to check it over himself before marking it.

  6. dumbcow
    • 4 years ago
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    ok answers match, just checked the books are factored a little more nicely, thats all

  7. anonymous
    • 4 years ago
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    Okay thanks. I'm stuck in between 2nd and 3rd derivative, where you do f''(r) = 4*3(4r+9)^2 + 12(4r+9)^2 + 96r(4r+9), where is that first term coming from?

  8. dumbcow
    • 4 years ago
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    for example: (4r+9)^3 + 12r(4r+9)^2 factor out a (4r+9)^2 (4r+9)^2(4r+9 +12r) =(16r+9) (4r+9)^2

  9. dumbcow
    • 4 years ago
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    oh right sorry i was afraid of that, it comes from derivative of first term of f'(r) -->(4r+9)^3 + 12r(4r+9)^2 ^

  10. anonymous
    • 4 years ago
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    Hmm, I see how you got the first f"(r) of the second derivative, but not how you got the 24(4r+9)^2 in the second step. I'll continue this from school, in about 1:30 hours. Wish me luck!

  11. dumbcow
    • 4 years ago
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    Ahh ...i just combined terms, 4*3=12 ...12+12 =24 good luck

  12. dumbcow
    • 4 years ago
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    oh and then i expanded 96r(4r+9)

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