f(x)=x^2/3 (x^2-4)...simplifiy then get f'(x)

• Mar 26th 2009, 12:25 PM
calcconfused
f(x)=x^2/3 (x^2-4)...simplifiy then get f'(x)
i know it simplifies to x^8/3 - 4x^2/3 but how does one get to that (esp. the x^8/3)

also, how do you get to f'(x) from here?

Thank you in advance to those smarter than I...
• Mar 26th 2009, 01:39 PM
sinewave85
Quote:

Originally Posted by calcconfused
i know it simplifies to x^8/3 - 4x^2/3 but how does one get to that (esp. the x^8/3)

also, how do you get to f'(x) from here?

Thank you in advance to those smarter than I...

Hi, calcconfused. To simplify the basic equation, remember that $\displaystyle x^{a}x^{b} = x^{a + b}$ and the rules for adding fractions:

$\displaystyle f(x) = x^{\frac{2}{3}}(x^{2} - 4)$

$\displaystyle f(x) = (x^{\frac{2}{3}})(x^{2}) - (x^{\frac{2}{3}})(4)$

$\displaystyle f(x) = (x^{\frac{2}{3}})(x^{\frac{6}{3}}) - (x^{\frac{2}{3}})(4)$

$\displaystyle f(x) = x^{\frac{8}{3}} - 4x^{\frac{2}{3}}$

And now the derivative:

$\displaystyle f^{\prime}(x) = \left(\frac{8}{3}\right)x^{(\frac{8}{3} - 1)} - \left(\frac{2}{3}\right)4x^{(\frac{2}{3} - 1)}$

$\displaystyle f^{\prime}(x) = \frac{8}{3}x^{\frac{5}{3}} - \frac{8}{3}x^{-\frac{1}{3}}$

$\displaystyle f^{\prime}(x) = \frac{8}{3}(x^{\frac{5}{3}} - x^{-\frac{1}{3}})$

Hope that helps!