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.
 A: Since $\frac{f^\prime(x)}{f(x)}=3\frac{a+b/x^2}{ax-b/x}$, the values of $f,\,f^\prime$ for $x=\tfrac32$ respectively give $a-b/x^2=\frac1x,\,a+b/x^2=\frac{10}{3}$, with solution $a=2,\,b=3$.
A: 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 $b=3$
A: Hint: From the conditions, we find
$$
\frac{{(9a - 4b)^3 }}{{72}} = 3 \; \text{and} \; \frac{{(9a + 4b)(9a - 4b)^2 }}{{36}} = 30.
$$
Multiplying both sides of the second equality with $(9a - 4b)^2$ and using the first, we get
$$
3(9a + 4b) = 15(9a - 4b),$$
i.e., $b = \frac{3}{2}a$. Can you proceed from here?
A: Well compute $f'$, and then plug in the two values, and solve the resulting system.
Let's see: $f'(x)=9(ax-b/x)^2(a+b/x^2)$.
Now $f(3/2)=3\implies 3(a(3/2)-b(2/3))=3$.  And $f'(3/2)=30\implies 9(a(3/2)-b(2/3))^2(a+b(4/9))=30$.
Thus you have two equations in two unknowns.
There's some algebra to be done.
