# How do you differentiate f(x)= 5/x^3 - 3/ x^(1/3)?

Oct 14, 2015

${f}^{'} \left(x\right) = - \frac{15}{x} ^ 4 + \frac{1}{x} ^ \left(\frac{4}{3}\right)$

#### Explanation:

Put it on root notation

$f \left(x\right) = 5 {x}^{- 3} - 3 {x}^{- \frac{1}{3}}$

Using the polynomial formula, ${f}^{'} \left(x\right) = n {x}^{n - 1}$, we have

${f}^{'} \left(x\right) = 5 \cdot \left(- 3\right) \cdot {x}^{- 3 - 1} - 3 \cdot \left(- \frac{1}{3}\right) \cdot {x}^{- \frac{1}{3} - 1}$

${f}^{'} \left(x\right) = - 15 {x}^{- 4} + {x}^{- \frac{4}{3}}$

Since the problem started with fraction notation, let us put the answer back in that notation

${f}^{'} \left(x\right) = - \frac{15}{x} ^ 4 + \frac{1}{x} ^ \left(\frac{4}{3}\right)$