For f(x)=sqrt [3]((64)/(x)) find f'(x) f'(x)=

To find the derivative of the function $f(x) = \sqrt[3]{\frac{64}{x}}$, we can use the chain rule and the power rule.<br /><br />First, let's rewrite the function in a more convenient form:<br />$f(x) = \left(\frac{64}{x}\right)^{1/3}$<br /><br />Now, we can differentiate this function using the chain rule:<br />$f'(x) = \frac{d}{dx}\left[\left(\frac{64}{x}\right)^{1/3}\right]$<br /><br />Using the chain rule, we have:<br />$f'(x) = \frac{1}{3}\left(\frac{64right)^{-2/3} \cdot \d}{dx}\left(\frac{64}{x}\right)$<br /><br />Now, let's differentiate $\frac{64}{x}$ with respect to $x$:<br />$\frac{d}{dx}\left(\frac{64}{x}\right) = -\frac{64}{x^2}$<br /><br />Substituting this back into our expression for $f'(x)$, we get:<br />$f'(x) = \frac{1}{3}\left(\frac{64}{x}\right)^{-2/3} \cdot \left(-\frac{64}{x^2}\right)$<br /><br />Simplifying this expression, we have:<br />$f'(x) = -\frac{64}{3x^2\left(\frac{64}{x}\right)^{2/3}}$<br /><br />Finally, we can simplify this expression further:<br />$f'(x) = -\frac{64}{3x^2\left(\frac{64}{x}\right)^{2/3}} = -\frac{64}{3x^2} \cdot \left(\frac{x}{64}\right)^{2/3} = -\frac{64}{3x^2} \cdot \frac{x^{2/3}}{64^{2/3}} = -\1}{3x^{4/3}}$<br /><br />Therefore, the derivative of the function $f(x) = \sqrt[3]{\frac{64}{x}}$ is:<br />$f'(x) = -\frac{1}{3x^{4/3}}$