Question

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)

Answer 1

Find f '' (x):

\[f(x) = \sqrt{2x^2+5}\]

Answer 2

Without simplifying it (not being sure if I can, also since I'm kinda lazy) I got a very messy answer of:

\[f '' (x) = \frac{ 8\sqrt{2x^2+5}-16x^2 }{ 8x^2+20[\sqrt{2x^2+5]} }\]

Answer 3

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

Answer 4

did you use the quotient rule or product rule for the second derv.

Answer 5

I used chain and quotient

Answer 6

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

Answer 7

Yeah I realize that but idk why I didn't do it that way, I just perfer the quotient in a situation like this

Answer 8

I'll do it like that one sec

Answer 9

I got something I don't wanna simplify lol

Answer 10

For the top I get
\[ 2 (2x^2+5)^{\frac{1}{2}} - x^2 (2x^2+5)^{-\frac{1}{2}} \]

Answer 11

and the bottom is
2x^2+5

Answer 12

With the chain and quotient I got

[4(2u^1/2) - 4x (u^1/2)] 4x all over (2u^1/2)^2

Answer 13

Since I didn't simplify f ' a first and kept

4x/2(2x^2+5)^1/2

Answer 14

the 2nd one would be ^1/2 in there

Answer 15

-1/2 jeez I cant even breathe right today

Answer 16

Using quotient rule, I got \(f''(x)=\dfrac{2\sqrt{2x^2+5}-4x^2(2x^2+5)^{-\frac{1}{2}}}{2x^2+5}\)

Answer 17

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)

Answer 18

I don't think you are doing the rule correctly, but I can't tell what you did.

Answer 19

I am writing it all out now and going to post a picture so you guys can see how I did it.

Answer 20

\[ \frac{2 (2x^2+5)^{\frac{1}{2}} (2x^2+5)^{\frac{1}{2}} - 4x^2 (2x^2+5)^{-\frac{1}{2}}(2x^2+5)^{\frac{1}{2}} }{(2x^2+5)^\frac{3}{2}} \\
=\frac{2(2x^2+5) - 4x^2}{(2x^2+5)^\frac{3}{2}} \\
= \frac{10}{(2x^2+5)^\frac{3}{2}} \]

Answer 21

used the chain rule and then the product rule.

Answer 22

Agreed?

Answer 23

yes

Answer 24

Then I plug u back in to get:

\[\frac{ 2 }{ (2x^2+5)^{1/2}} - \frac{ 4x^2 }{ (2x^2+5)^{3/2} }\]

Agreed?

Answer 25

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.

Answer 26

yes, and if you multiply the first term by u/u it simplifies

Answer 27

your common denominator will be \( (2x^2+5)^\frac{3}{2} \)

Answer 28

or least, that will be the LCD

Answer 29

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

Answer 30

that is messy. use (stuff)^(3/2)

Answer 31

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

Answer 32

Simplifying is my enemy, honestly.

Answer 33

you could and it's not wrong. but notice
\[ \frac{ 2(2x^2+5) }{ (2x^2+5)^{1/2}(2x^2+5) } - \frac{ 4x^2 }{ (2x^2+5)^{3/2} } \]

Answer 34

You have an extra (2x^2+5)

Answer 35

I multiplied by top and bottom

Answer 36

?

Answer 37

Oh I see what you're doing.

Answer 38

You're doing that so you'll just have (2x^2+5)^3/2

Answer 39

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

Answer 40

10/(2x^2+5)^2/3

Answer 41

^3/2

but yes, that looks a little better.

Answer 42

I see what you did, yeah that's a lot easier. I just suck at simplifying lol

Answer 43

It is like a lot of things. Practice.

Answer 44

Yeah definitely agree lol

Answer 45

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