Square root - calculus (1 Viewer)

[sando]

i'm having trouble with this question so i thought i'd post it up:

Find the exact value of f"(2) if f(x) = x sqrt (3x-4)

 

[Mountain.Dew]

i'm having trouble with this question so i thought i'd post it up:
Find the exact value of f"(2) if f(x) = x sqrt (3x-4)

sando, r u familiar with the product rule?

split f(x) into u = x, v = sqrt (3x-4)
then, find u' and v'. note that for v' u need to use the chain rule.

then, f'(x) = uv' + vu'
then u can find f'(2)

 

[sando]

i need to find f"(x) not f'(x)

and yes i am aware of the product rule and chain rule but i jsut can't seem to get the correct answer

 

[pLuvia]

For your question

f'(x)=3x/2*(3x-4)^-1/2 + (3x-4)^1/2
f''(x)= -9x/4*(3x-4)^-3/2 + 3*(3x-4)^-1/2
Now just sub x = 2 and you'll get your answer

Edit: I did it for you, is the answer 3/4sqrt2 ?

 

[sando]

thanks... i managed to find f'(x) but not f"(x)

maybe i better revise the rules

 

[sando]

yep, ur right

the answer is:

3 / 4sqrt2 or 3sqrt2 / 8

 

[sando]

damn't.... wat is wrong with me today

can someone plz care to post up a full worked solution

thanks

 

[shinji]

f"(x) is just the derivative of f'(x)

to do this question; u need to know the indice law.

 

[sando]

ok, let me clear this up, i know how to do f'(x) and f"(x) sums and i know the laws

i found the same f"(x) answer as plivia:

f''(x)= -9x/4*(3x-4)-3/2 + 3*(3x-4)-1/2

however, i cannot workout the next step, cos u can't simply sub 2 into that.

can someone write it in a simpler form plz????

 

[SoulSearcher]

ok, let me clear this up, i know how to do f'(x) and f"(x) sums and i know the laws
i found the same f"(x) answer as plivia:
f''(x)= -9x/4*(3x-4)^-3/2 + 3*(3x-4)^-1/2
however, i cannot workout the next step, cos u can't simply sub 2 into that.
can someone write it in a simpler form plz????

you can substitute 2 into that, you just have to rearrange the question to make it easier

f''(x) = -9x/4*(3x-4)^-3/2 + 3*(3x-4)^-1/2
f''(x) = 3*(3x-4)^-1/2 -9x/4*(3x-4)^-3/2
f''(x) = 3 / sqrt (3x-4) - 9x / 4 sqrt (3x-4)³
therefore
f''(2) = 3 / sqrt (6-4) - 9*2 / 4 sqrt (6-4)³
f''(2) = 3 / sqrt 2 - 18 / 4 sqrt 8
f''(2) = 3 sqrt 2 / 2 - 18 sqrt 8 / 32
f''(2) = { 48 sqrt 2 - 18 * 2 sqrt 2 } / 32
f''(2) = { 48 sqrt 2 - 36 sqrt 2 } / 32
f''(2) = 12 sqrt 2 / 32
f''(2) = 3 sqrt 2 / 8

 

[sando]

thanx so much, this question was troubling me all day

i can't believe that i got up to this line:

-9x/4*(3x-4)-3/2 + 3*(3x-4)-1/2


and failed to regonise what to do

 

[Mountain.Dew]

i need to find f"(x) not f'(x)
and yes i am aware of the product rule and chain rule but i jsut can't seem to get the correct answer

ahhhhhhhh the fonts fault...the f ' ' (x) looked a lot like f ' (x) when there are no spaces

f''(x) as opposed to f'(x)...they looked pretty similar, so i couldnt tell if u where asking for f ' ' (2) or f ' (2)

my bad, sorry, need change my browser text font.

 

[Riviet]

If you're a firefox user, you can press Ctrl and + or - keys to zoom in and out. It should enhance your experience on bos forums.

 

[sando]

what the hell is firefox?

 

[Riviet]

Firefox is a web browser, click here for more info. I personally love it because you can have multiple tabs rather than several
explorer windows.

 

[Mountain.Dew]

If you're a firefox user, you can press Ctrl and + or - keys to zoom in and out. It should enhance your experience on bos forums.

thank you very much riviet! *bows to riviet*