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?
 A: $$f^3 (x)=x^2 (6-x) $$
by differentiation
$$3f^2 (x)f'(x)=2x (6-x)-x^2=3x (4-x) $$
thus
$$f'(x)=\frac {x (4-x)}{f^2 (x)} $$
$$=\frac {x (4-x)}{x^\frac43 (6-x)^\frac23} $$
$$=\frac {4-x}{(6-x)^\frac23}x^{\frac {-1}{3}}$$
A: 
Pulling out this common factor is fine. We can simplify the last expression slightly more and obtain
  \begin{align*}
\color{blue}{-x^{\frac{2}{3}}}&\color{blue}{ \cdot \frac{1}{3} (6-x) ^{-\frac{2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {-\frac{1}{3}}}\\
&=\frac{1}{3}x^{-\frac{1}{3}}(6-x)^{-\frac{2}{3}}(-x+(6-x)2)\\
&=\frac{1}{3}x^{-\frac{1}{3}}(6-x)^{-\frac{2}{3}}(12-3x)\\
&\,\,\color{blue}{=x^{-\frac{1}{3}}(6-x)^{-\frac{2}{3}}(4-x)}
\end{align*}

A: We've got to preserve the terms here, and we see that the ordering is mismatched, as in 
$$\frac{d}{dx}(x^{\frac{2}{3}}(6-x)^{\frac{2}{3}})=\frac{2}{3}(\frac{6-x}{x})^{\frac{1}{3}}-\frac{1}{3}(\frac{x}{6-x})^{\frac{2}{3}}$$
Then we can factor out $(\frac{6-x}{x})^{\frac{1}{3}}$ to make the second term much nicer with no powers
Your answer and this one are ultimately the same
A: Alternatively:
$$\frac{d}{dx}\left(x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}\right)=\frac{d}{dx}\left((6x^2-x^3)^{\frac13}\right)=\frac{1}{3}(6x^2-x^3)^{-\frac{2}{3}}(12x-3x^2)=\\
\frac{4x-x^2}{(6x^2-x^3)^{\frac{2}{3}}}=\frac{4x-x^2}{x^{\frac{4}{3}}(6-x)^{\frac23}}=\frac{4-x}{x^{\frac{1}{3}}(6-x)^{\frac23}}.$$
