# calculus please, higher order derivatives

• Nov 28th 2006, 05:52 AM
amanda1410
Neem some urgent help on these please, if possible can you show me the method also so I can work out simular ones please

http://img.photobucket.com/albums/v1...untitled-3.jpg
• Nov 28th 2006, 05:59 AM
ThePerfectHacker
Quote:

Originally Posted by amanda1410
Neem some urgent help on these please, if possible can you show me the method also so I can work out simular ones please

http://img.photobucket.com/albums/v1...untitled-3.jpg

I will do,
$\displaystyle F'''(x)$

Do you see how to take derivative?
You need derivative of,
$\displaystyle F''(x)=-2x(1+x^2)^{-2}$
Use product rule,
$\displaystyle (-2x)'(1+x^2)^{-2}+(-2x)[(1+x^2)^{-2}]'$
Thus, use chain rule on the second one,
$\displaystyle (-2)(1+x^2)^{-2}+(-2x)(2x)(-2)(1+x^2)^{-3}$
Thus,
$\displaystyle -2(1+x^2)^{-2}-8x^2(1+x^2)^{-3}$
Thus,
$\displaystyle -2(1+x^2)^{-2}[1+4x(1+x^2)^{-1}]$
• Nov 28th 2006, 10:18 AM
Soroban
Hello, Amanda!

This one of the more unpleasant functions to differentiate.
I suggest using the Quotient Rule.
. . Those negative exponents can be dangerous.

It starts simply enough . . .

$\displaystyle f(x)\;=\;\tan^{-1}(x)$

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

$\displaystyle f''(x)\;=\;-1(x^2+1)^{-1}\cdot2x \;=\;-2\,\frac{x}{x^2+1}$

. . If there is a coefficient, I "pull it out front".

Now it gets very messy . . . simplify each derivative before continuing.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle f'''(x)\;=\;-2\left[\frac{(x^2+1)\cdot1 - x\cdot2(x^2+1)\cdot2x}{(x^2+1)^4}\right]$ $\displaystyle \;=\;-2\left[\frac{(x^2+1)^2 - 4x^2(x^2+1)}{(x^2+1)^4}\right]$

Factor: .$\displaystyle -2\,\frac{(x^2+1)\left(x^2+1 - 4x^2\right)}{(x^2+1)^4} \;= \;-2\,\frac{(1 - 3x^2)}{(x^2+1)^3} \;= \;2\,\frac{3x^2-1}{(x^2+1)^3}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

$\displaystyle f^{(4)}(x) \;=\;2\,\left[\frac{(x^2+1)^3\cdot6x - (3x^2-1)\cdot3(x^2+1)^2\cdot2x}{(x^2+1)^6}\right]$

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

Factor: .$\displaystyle 2\cdot6x(x^2+1)^2\left[\frac{(x^2+1) - (3x^2 - 1)}{(x^2+1)^6}\right] \;= \;12x\,\frac{2-2x^2}{(x^2+1)^4}\;= \;24\,\frac{x-x^3}{(x^2+1)^4}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

There were at least $\displaystyle 10^{37}$ opportunities to make errors.
. . I don't guarentee any of my results.

• Jan 23rd 2007, 03:04 PM
Quote:

Originally Posted by Soroban
[size=3]Hello, Amanda!

This one of the more unpleasant functions to differentiate.
I suggest using the Quotient Rule.
. . Those negative exponents can be dangerous.

It starts simply enough . . .

$\displaystyle f(x)\;=\;\tan^{-1}(x)$

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

$\displaystyle f''(x)\;=\;-1(x^2+1)^{-1}\cdot2x \;=\;-2\,\frac{x}{x^2+1}$

i think its $\displaystyle -(x^2+1)^{-2}\cdot2x$? is it?