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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.
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I see you typing dumbcow. You're awesome :)

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0this 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

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0depending on how the textbook simplifies answers, it might be same thing but look different is everything expanded out in their answers?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here'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(8x3))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah 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.

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0ok answers match, just checked the books are factored a little more nicely, thats all

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay 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?

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0for 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

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0oh right sorry i was afraid of that, it comes from derivative of first term of f'(r) >(4r+9)^3 + 12r(4r+9)^2 ^

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm, 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!

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0Ahh ...i just combined terms, 4*3=12 ...12+12 =24 good luck

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0oh and then i expanded 96r(4r+9)
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