anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
I see you typing dumbcow. You're awesome :)
dumbcow
  • dumbcow
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
dumbcow
  • dumbcow
depending on how the textbook simplifies answers, it might be same thing but look different is everything expanded out in their answers?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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))
anonymous
  • anonymous
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.
dumbcow
  • dumbcow
ok answers match, just checked the books are factored a little more nicely, thats all
anonymous
  • anonymous
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?
dumbcow
  • dumbcow
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
dumbcow
  • dumbcow
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 ^
anonymous
  • anonymous
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!
dumbcow
  • dumbcow
Ahh ...i just combined terms, 4*3=12 ...12+12 =24 good luck
dumbcow
  • dumbcow
oh and then i expanded 96r(4r+9)

Looking for something else?

Not the answer you are looking for? Search for more explanations.