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