# Thread: Derivative

1. ## Derivative

By using the power rule, how can i find the derivative of:

(x-1)^2 *x / x^1/3

I just dont know how to do the numerator..

should i take the 2 infront and then simplify or sould i resolve the power, into (x-1)(x-1) ???

2. Originally Posted by HHHH
By using the power rule, how can i find the derivative of:

(x-1)^2 *x / x^1/3

I just dont know how to do the numerator..

should i take the 2 infront and then simplify or sould i resolve the power, into (x-1)(x-1) ???
note that $\frac{x}{x^{\frac{1}{3}}} = x^{\frac{2}{3}}$

$\frac{d}{dx} \left[(x-1)^2 \cdot x^{\frac{2}{3}} \right]$

product rule ...

$(x-1)^2 \cdot \frac{2}{3}x^{-\frac{1}{3}} + x^{\frac{2}{3}} \cdot 2(x-1)$

3. Originally Posted by skeeter
note that $\frac{x}{x^{\frac{1}{3}}} = x^{\frac{2}{3}}$

$\frac{d}{dx} \left[(x-1)^2 \cdot x^{\frac{2}{3}} \right]$

product rule ...

$(x-1)^2 \cdot \frac{2}{3}x^{-\frac{1}{3}} + x^{\frac{2}{3}} \cdot 2(x-1)$

thanx, but how or why did you do the last part, +x^2/3 * 2(x-1) ?

4. Originally Posted by HHHH
thanx, but how or why did you do the last part, +x^2/3 * 2(x-1) ?
He held $x^{\frac{2}{3}}$ constant and took the derivative of $(x-1)^2$, which is 2(x-1). This is how the product rule works. d(ab)=a'b+b'a