Differentiating a a rational function

[Specter]

Homework Statement

Find the first and second derivatives of $\displaystyle f(x)=\frac {1} {x^2+6}$

Homework Equations

The Attempt at a Solution

[/B]
$\displaystyle f(x)=\frac {1} {x^2+6}$

$\displaystyle f(x)=(x^2+6)^{-1}$

$\displaystyle f'(x)=-1(2x)(x^2+6)^{-2}$

$\displaystyle =-2x(x^2+6)^{-2}$

$\displaystyle =-\frac {2x} {(x^2+6)^2}$

I am getting an incorrect answer for the second derivative.

$\displaystyle f'(x)=-\frac {2x} {(x^2+6)^2}$

$\displaystyle f'(x)=-2x(x^2+6)^{-2}$

Following the chain rule..

$\displaystyle F''(x)=nf'(x)f(x)^{n-1}$

$\displaystyle F''(x)=-2x(-2)(2x)(x^2+6)^{-3}$

$\displaystyle =8x^2(x^2+6)^{-3}$

$\displaystyle =\frac {8x^2} {(x^2+6)^3}$

The second derivative is supposed to be $\displaystyle f''(x)=\frac {6x^2-12} {(x^2+6)^3}$ . I can't find my mistake, I thought that I used the chain rule correctly.

 

[Orodruin]

I thought that I used the chain rule correctly.

You did not. There is no chain rule for the second derivative. You need to apply the product rule for derivatives when you differentiate the first derivative to get the second.

 

[Specter]

You did not. There is no chain rule for the second derivative. You need to apply the product rule for derivatives when you differentiate the first derivative to get the second.

Ohhhh.

$\displaystyle f'(x)=-2x(x^2+6)^{-2}$

$\displaystyle f''(x)=-2(x^2+6)^{-2}+(-2x)(-2)(x^2+6)^{-3}(2x)$

$\displaystyle =\frac {-2} {(x^2+6)^2} + \frac {8x^2} {(x^2+6)^3}$

$\displaystyle = \frac {-2(x^2+6)+8x^2} {(x^2+6)^3}$

$\displaystyle = \frac {6x^2-12} {(x^2+6)^3}$

Thank you!

 

[MidgetDwarf]

Why not apply the quotient rule directly?