AP Calc BC help! I recently took my test and I'm doing my test corrections. I originally got this problem wrong and I was very surprised. If someone could check my new answer, I'd greatly appreciate it. (Warning it's a somewhat lengthy problem)
Find f '' (x): \[f(x) = \sqrt{2x^2+5}\]
Without simplifying it (not being sure if I can, also since I'm kinda lazy) I got a very messy answer of: \[f '' (x) = \frac{ 8\sqrt{2x^2+5}-16x^2 }{ 8x^2+20[\sqrt{2x^2+5]} }\]
I know my first derivative is right because that's just a basic chain rule. But the 2nd is maybe iffy. f '(x)= 2x/(2x^2+5)^1/2
did you use the quotient rule or product rule for the second derv.
I used chain and quotient
To make second derivative easier, you can just do this: \(f'(x) = \dfrac{2x}{\sqrt{2x^2+5}}\Rightarrow f'(x) = 2x(2x^2+5)^{-\frac{1}{2}}\) Use product rule from here
Yeah I realize that but idk why I didn't do it that way, I just perfer the quotient in a situation like this
I'll do it like that one sec
I got something I don't wanna simplify lol
For the top I get \[ 2 (2x^2+5)^{\frac{1}{2}} - x^2 (2x^2+5)^{-\frac{1}{2}} \]
and the bottom is 2x^2+5
With the chain and quotient I got [4(2u^1/2) - 4x (u^1/2)] 4x all over (2u^1/2)^2
Since I didn't simplify f ' a first and kept 4x/2(2x^2+5)^1/2
the 2nd one would be ^1/2 in there
-1/2 jeez I cant even breathe right today
Using quotient rule, I got \(f''(x)=\dfrac{2\sqrt{2x^2+5}-4x^2(2x^2+5)^{-\frac{1}{2}}}{2x^2+5}\)
d u/v = (v du - u dv)/v^2 in this case u = 2x du = 2 v= sqr(2x^2+5) dv = 2x/sqr(2x^2+5)
I don't think you are doing the rule correctly, but I can't tell what you did.
I am writing it all out now and going to post a picture so you guys can see how I did it.
\[ \frac{2 (2x^2+5)^{\frac{1}{2}} (2x^2+5)^{\frac{1}{2}} - 4x^2 (2x^2+5)^{-\frac{1}{2}}(2x^2+5)^{\frac{1}{2}} }{(2x^2+5)^\frac{3}{2}} \\ =\frac{2(2x^2+5) - 4x^2}{(2x^2+5)^\frac{3}{2}} \\ = \frac{10}{(2x^2+5)^\frac{3}{2}} \]
used the chain rule and then the product rule.
Agreed?
yes
Then I plug u back in to get: \[\frac{ 2 }{ (2x^2+5)^{1/2}} - \frac{ 4x^2 }{ (2x^2+5)^{3/2} }\] Agreed?
Then I get a common denominator by doing (2x^2+5)^1/2 times (2x^2+5)^3/2 Which gives me (2x^2+5)^2 as my denominator.
yes, and if you multiply the first term by u/u it simplifies
your common denominator will be \( (2x^2+5)^\frac{3}{2} \)
or least, that will be the LCD
No it's (2x^2+5)^2 because you multiply the 2 denominators together for a common and 1/2 +3/2 = 4/2 which is 2
that is messy. use (stuff)^(3/2)
And I muliply the left by (2x^2+5)^3/2 and the right by (2x^2+5)^1/2 To get a final answer of: 2(2x^2+5)^3/2 - 4x^2(2x^2+5)^1/2 all over (2x^2+5)^2
Simplifying is my enemy, honestly.
you could and it's not wrong. but notice \[ \frac{ 2(2x^2+5) }{ (2x^2+5)^{1/2}(2x^2+5) } - \frac{ 4x^2 }{ (2x^2+5)^{3/2} } \]
You have an extra (2x^2+5)
I multiplied by top and bottom
?
Oh I see what you're doing.
You're doing that so you'll just have (2x^2+5)^3/2
This was your result \[ \frac{ 2 }{ (2x^2+5)^{1/2}} - \frac{ 4x^2 }{ (2x^2+5)^{3/2} } \] multiply the first term now you can simplify the top
10/(2x^2+5)^2/3
^3/2 but yes, that looks a little better.
I see what you did, yeah that's a lot easier. I just suck at simplifying lol
It is like a lot of things. Practice.
Yeah definitely agree lol
Thanks so much for helping, I have a couple others in a minute that are different so if you see me post maybe you can help again lol
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