# PracticeDifferentiate 1

Differentiate.

#### Jason

Staff member
Moderator
$$\displaystyle y = \sqrt[3]{x^{2}} * (2x - x^{2})$$

Without product rule:

$$\displaystyle y = x^{2/3}(2x - x^{2})$$

$$\displaystyle y = 2x^{5/3} - x^{8/3}$$

$$\displaystyle y' = \dfrac{5}{3}x^{2/3} - \dfrac{8}{3}x^{5/3}$$

Last edited:

#### MarkFL

##### La Villa Strangiato
Staff member
Moderator
Math Helper
Jason, the first term on the RHS of your final result has the original coefficient of 2 left off.

#### Jason

Staff member
Moderator
Jason, the first term on the RHS of your final result has the original coefficient of 2 left off.
$$\displaystyle y = \sqrt[3]{x^{2}} * (2x - x^{2})$$

Without product rule:

$$\displaystyle y = x^{2/3}(2x - x^{2})$$

$$\displaystyle y = 2x^{5/3} - x^{8/3}$$

$$\displaystyle y' = 2(\dfrac{5}{3})x^{2/3} - \dfrac{8}{3}x^{5/3}$$

$$\displaystyle y' = \dfrac{10}{3}x^{2/3} - \dfrac{8}{3}x^{5/3}$$

harpazo and MarkFL

Good job.