Derivative

**HHHH** · November 23rd 2009, 12:28 PM

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) ???

**skeeter** · November 23rd 2009, 01:18 PM

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 ...

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

**HHHH** · November 24th 2009, 12:20 PM

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) ?

**Jameson** · November 24th 2009, 12:45 PM

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

Thread: Derivative

LinkBack

Thread Tools

Display

Derivative

Similar Math Help Forum Discussions

contuous weak derivative $\Rightarrow$ classic derivative ?

Expanding a time derivative in terms of other variables /// Material derivative

[SOLVED] Definition of Derivative/Alt. form of the derivative

Derivative of arctan in a partial derivative

derivative of 1/[sqrt[x]: proving with derivative definition

Search Tags