Finding constants $a$ and $b$ using derivatives 'Suppose $f(x)= 3(ax-b/x)^3$. Given that $f(3/2)= 3$ and $f'(3/2)= 30$, find $a$ and $b$.'
I've tried chain rule and getting $a$ or $b$ on its own and substituting back into the function, but I feel like I'm overcomplicating it. 
Thanks for any help.

We have $f'(x)=9(a+\frac{b}{x^2})(ax-\frac{b}{x})^3$ 
$f(\frac{3}{2})=3$ so $3(a.\frac{3}{2}-b\frac{2}{3})^3=3$; $f'(\frac{3}{2})=30$ so $9(a+b\frac{4}{9})(a\frac{3}{2}-b\frac{2}{3})^3$
Implies $9a-4b=6$ and $9a+4b=30$, so $a=2$ and $b=3$
