Simplifying the derivative of $x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$ $$x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$$
So I get:
$$-x^{\frac{2}{3}} \cdot \frac{1}{3} (6-x) ^{\frac{-2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {\frac{-1}{3}}$$
How does one go about simplifying this?
I guess I can pull out common terms like this:
$$\frac{1}{3} x ^{-\frac{1}{3}} (6-x)^{\frac{-2}{3}} ( -x + (6-x) \cdot 2)$$
Is that right?

$$f^3 (x)=x^2 (6-x) $$
by differentiation
$$3f^2 (x)f'(x)=2x (6-x)-x^2=3x (4-x) $$
thus
$$f'(x)=\frac {x (4-x)}{f^2 (x)} $$
$$=\frac {x (4-x)}{x^\frac43 (6-x)^\frac23} $$
$$=\frac {4-x}{(6-x)^\frac23}x^{\frac {-1}{3}}$$
