If f(2)=6, f'(2)=-4, and f"(2)=2, what is (d^2*(f^3 (x)))/(dx^2 ) at x=2?

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- anonymous

If f(2)=6, f'(2)=-4, and f"(2)=2, what is (d^2*(f^3 (x)))/(dx^2 ) at x=2?

- schrodinger

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- myininaya

\[\frac{d^2([f(x)]^3)}{dx^2}\]
\[=\frac{d}{dx}(3[f(x)]^2f'(x))=3[2f(x)f'(x)f'(x)+[f(x)]^2f''(x)]\]

- myininaya

\[3[f(2)f'(2)f'(2)+[f(2)]^2f''(2)]\]

- myininaya

now plug in and evaluate and simplify

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## More answers

- anonymous

Thank you I'm going to try it now.

- myininaya

I used chain rule and product rule by the way

- anonymous

okay :)

- anonymous

The answers I keep coming up don't seem correct. Can you please explain in more detail.

- myininaya

lol and I didn't notice you were having problems with this money sorry

- amistre64

just like a woman, they state their opinions and then yougotta live with em ;)

- myininaya

\[3[6(-4)(-4)+(6)^2(2)]=3[6(16)+36(2)]=3[96+72]=3[168]\]

- myininaya

= 504 is what I'm getting

- amistre64

ipso facto

- anonymous

I was missing the 2 in front of the f(x). But I understand now.

- amistre64

i dont think id of come to that today;

- myininaya

lol poor amistre
do you ever come to it? ;)

- anonymous

The radius of a sphere is increasing at a constant rate of 2cm/sec. At the instant when the volume of the sphere is 36π cm^3, what is the rate that the surface area is increasing? Surface area=4π r^2 Volume=4/3π r^3 This one I need a full full explanation on

- myininaya

\[V(t)=\frac{4}{3} \pi [r(t)]^3=> V'(t)=\frac{4}{3} \pi 3 [r(t)]^2 r'(t)\]

- myininaya

\[\S(t)=4 \pi [r(t)]^2 => \S'(t)=4 \pi 2r(t) r'(t)=8 \pi r(t)r'(t)\]

- myininaya

\[V'(t)=4 [r(t)]^2 r'(t)\]

- myininaya

so r'=2 do you see that in the first sentence?

- myininaya

\[V=36 \pi\] from second sentence

- myininaya

we want to know S' based on second sentence

- myininaya

\[36=\frac{4}{3} \pi r^3\]
solve this for r

- myininaya

to find S' this is the only thing we need since r' is already given

- myininaya

oops the V=36 pi

- myininaya

\[36 \pi =\frac{4}{3} \pi r^3\] *

- myininaya

solve that for r

- anonymous

So am I suppose to divide both sides by 4/3\[\pi\] ?

- myininaya

or multiply both sides by 3/(4pi)

- anonymous

so would I have 27pi=r^3?

- myininaya

well the pi's would cancel

- anonymous

27=r^3?

- myininaya

yes since pi/pi=1
so r=3

- myininaya

\[\S'=8 \pi r r'\]

- myininaya

and remember r'=2

- myininaya

So now you can find S'

- anonymous

So I should come up with 48pi cm^2/sec correct?

- myininaya

48 pi is right!

- anonymous

Thanks!!

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