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
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
What I'm saying is I think that this way is the easiest. However, if you use th equotient rule then
$\displaystyle
\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} $$\displaystyle =\frac{ 9(9 - x^2)}{x^4} $