1. ## Simplify derivative

Can someone please show me how this derivative: f'(x) = x^3(18x) - (9)(x^2 - 3)(3x^2)/(x^3)^2

Simplifies to this: 9(9 - x^2)/ x^4

Because I am lost lol

Here is the original equation:
f(x)= 9(x^2-3)/x^3

Thanks

2. Originally Posted by Jwd41190
Can someone please show me how this derivative: f'(x) = x^3(18x) - (9)(x^2 - 3)(3x^2)/(x^3)^2

Simplifies to this: 9(9 - x^2)/ x^4

Because I am lost lol

Here is the original equation: f(x)= 9(x^2-3)/x^3

Thanks
It's really easier if you expand

$
f(x) = \frac{9x^2}{x^3} - \frac{27}{x^3} = \frac{9}{x} - \frac{27}{x^3}
$

3. Thats still not right is it here it is from the book

Here it is from the book:
http://img121.imageshack.us/img121/3689/scn0001.jpg

4. Originally Posted by Danny
It's really easier if you expand

$
f(x) = \frac{9x^2}{x^3} - \frac{27}{x^3} = \frac{9}{x} - \frac{27}{x^3}
$
What I'm saying is I think that this way is the easiest. However, if you use th equotient rule then

$
\frac{x^3 (18x) - 9(x^2-3)(3x^2)}{x^6} = \frac{18x^4 - 27x^4+81x^2}{x^6} = \frac{ - 9x^4+81x^2}{x^6} =\frac{ - 9x^2+81}{x^4}$
$=\frac{ 9(9 - x^2)}{x^4}$