Question

f' and f'' of : f(x)= (x^3)/3 - 1/x ??

Answer 1

Okay, so first we re-write it: f(x)= (x^3)/3 - 1/x
therefore f(x) = (x^3)/3 - x^-1
tf, following the rules of differentiation, f'(x) = (3x^2)/3 - (-1)x^-2...
tf f'(x) = x^2 + x^-2
tf f'(x) = x^2 + 1/x^2

Answer 2

Now, copying what I did there, what would f''(x) be...?

Answer 3

yes I did the first part, in fact, I have difficulties with the other part...

Answer 4

(f'')

Answer 5

That'S what I did but it's not the goood answer

Answer 6

Okay, I see what happened. You are correct, to a point. xD
f"(x) does indeed equal 2x - 2x^-3
but then the next step is that f"(x) = 2x-2/x^3, not (2x-2)/x^3.

Answer 7

whats the difference?

Answer 8

the real answer is :( 2(x^2 + 1)(x-1)(x+1) ) / x^3

Answer 9

O.o

Answer 10

I'm flummoxed. LOL

Answer 11

But let me think...

Answer 12

OHHHH

Answer 13

I have no idea why they wrote it like this, but if we expand 2(x^2 + 1)(x-1)(x+1) ) / x^3, we eventually come out at 2x-2/x^3. LOL

Answer 14

okay haha thanls :P