anonymous
  • anonymous
find the derivative and second derivative of (x^2-9)^2
Calculus1
  • 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.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amoodarya
  • amoodarya
\[y=(x^2-9)^2=\\x^4-18x^2+81\\\]it is easier now to find f',f''
anonymous
  • anonymous
Wouldn't it just be a chain rule from step 1?
amoodarya
  • amoodarya
|dw:1437519674083:dw|

Looking for something else?

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

More answers

amoodarya
  • amoodarya
|dw:1437519721063:dw|
anonymous
  • anonymous
That's too confusing lol could you just do the chain rule?
triciaal
  • triciaal
|dw:1437519701313:dw|
anonymous
  • anonymous
Thank you. What about f''(x)?
amoodarya
  • amoodarya
that was chain rule
anonymous
  • anonymous
Sorry amood it was just your hand writing haha no offense
amoodarya
  • amoodarya
what 's f'= ? simplify and writ it here plz
anonymous
  • anonymous
\[4x(x^2-9)\]
triciaal
  • triciaal
|dw:1437519790039:dw|
amoodarya
  • amoodarya
distribute 4x in (x^2-9 ) then take y''
anonymous
  • anonymous
Done, thank you sir Triciaal.
anonymous
  • anonymous
Thanks Amood
triciaal
  • triciaal
@MBilbeisi either way you will get the same result
triciaal
  • triciaal
I am not a Sir
anonymous
  • anonymous
@Triciaal are you good at extreme values?
anonymous
  • anonymous
Sorry, madam.
triciaal
  • triciaal
ok
anonymous
  • anonymous
You are?
triciaal
  • triciaal
maximum and minimum
anonymous
  • anonymous
Yes!!
anonymous
  • anonymous
It's a really hard question i can't seem to get it right
anonymous
  • anonymous
Maybe not so hard for you
triciaal
  • triciaal
when 2nd derivative at critical point is negative you have max when positive then you have minimum
triciaal
  • triciaal
the 2nd derivative is showing the change in the slope the slope is the 1st derivative
anonymous
  • anonymous
So if the question says Find the extreme values of the function and where they occur, i find f'(x) and f''(x)?
triciaal
  • triciaal
out of practice but I had learnt it well enough
anonymous
  • anonymous
\[\left(\begin{matrix}1 \\ x^2-1\end{matrix}\right)\]
triciaal
  • triciaal
find the critical points use in 2nd derivative
triciaal
  • triciaal
sorry have to leave but post question will check back later if still need help
anonymous
  • anonymous
Okay i got you, so for that function i do f', by quotient rule, and then critical values, then f''?
anonymous
  • anonymous
okay i will, thanks
triciaal
  • triciaal
x not = 1 or -1 for this because do not want to divide by 0

Looking for something else?

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