could some please give me the answer for f'(x), f''(x) and f'''(x) for this function please, i'm tryin to do it but the page is gettin so messy, and im gonna pull my hair out if i try for any longer!!

the function is
$\displaystyle f(x) = (x^3+3x^2+1)^{\frac{1}{3}}$

2. Originally Posted by mathmonster
could some please give me the answer for f'(x), f''(x) and f'''(x) for this function please, i'm tryin to do it but the page is gettin so messy, and im gonna pull my hair out if i try for any longer!!

the function is
$\displaystyle f(x) = (x^3+3x^2+1)^{\frac{1}{3}}$
$\displaystyle f'(x) = \frac{1}{3} (x^3+3x^2+1)^{\frac{-2}{3}} (3x^2 + 6x)$

$\displaystyle f''(x) = \frac{1}{3} \left[ {(x^3+3x^2+1)^{\frac{-2}{3}} (6x + 6) + \frac{-2}{3}(x^3 + 3x^2 + 1)^{\frac{-5}{3}} (3x^2 + 6x)^2} \right]$

$\displaystyle f'''(x) = \frac{1}{3} \left[ {6(x^3+3x^2+1)^{\frac{-2}{3}} + (6x+6)\frac{-2}{3}(x^3 + 3x^2 + 1)^{\frac{-5}{3}} (3x^2 + 6x) }\right]$

$\displaystyle + \frac{1}{3} \left[ { \frac{-4}{3}(x^3 + 3x^2 + 1)^{\frac{-5}{3}} (3x^2 + 6x)(6x+6) }\right]$

$\displaystyle + \frac{1}{3} \left[ { \frac{10}{9} (x^3 + 3x^2 + 1)^{\frac{-8}{3}} (3x^2 + 6x)^3} \right]$

just simplify it if you want..
i hope i got the final answer right.. Ü

3. Originally Posted by mathmonster
could some please give me the answer for f'(x), f''(x) and f'''(x) for this function please, i'm tryin to do it but the page is gettin so messy, and im gonna pull my hair out if i try for any longer!!

the function is
$\displaystyle f(x) = (x^3+3x^2+1)^{\frac{1}{3}}$
$\displaystyle f'(x) = \frac{x^2+2x}{f^2(x)}$

Then:

$\displaystyle f''(x)=\frac{2(x+1)}{f^2(x)} + \frac{(-2)f'(x)(x^2+2)}{f^3(x)}$

$\displaystyle =\frac{2}{f^2(x)}\left[(x+1)-\frac{f'(x)(x^2+2}{f(x)} \right]$

and so on.

RonL