# How do you find the derivative of f(x)=(5x^6sqrt x) + (3/(x^3 sqrt x))?

##### 1 Answer
Jul 22, 2016

$f ' \left(x\right) = \frac{1}{2} \left(65 {x}^{5} \sqrt{x} - \frac{21}{{x}^{4} \sqrt{x}}\right)$

#### Explanation:

$f \left(x\right) = \left(5 {x}^{6} \sqrt{x}\right) + \left(\frac{3}{{x}^{3} \sqrt{x}}\right)$

Using the rules of indicies $f \left(x\right)$ can be written:

$f \left(x\right) = 5 {x}^{6} {x}^{\frac{1}{2}} + 3 {x}^{-} 3 {x}^{- \frac{1}{2}}$ $= 5 {x}^{\frac{13}{2}} + 3 {x}^{- \frac{7}{2}}$

Aplying the Power Rule to both terms:

$f ' \left(x\right) = 5 \cdot \frac{13}{2} {x}^{\frac{13}{2} - 1} + 3 \cdot \left(- \frac{7}{2}\right) {x}^{- \frac{7}{2} - 1}$
$= \frac{1}{2} \left(65 {x}^{\frac{11}{2}} - 21 {x}^{- \frac{9}{2}}\right)$

To express $f ' \left(x\right)$ in the form of $f \left(x\right)$ in the original question, we can rewrite $f ' \left(x\right)$ as:
$f ' \left(x\right) = \frac{1}{2} \left(65 {x}^{5} \cdot {x}^{\frac{1}{2}} - 21 {x}^{- 4} \cdot {x}^{- \frac{1}{2}}\right)$

$= \frac{1}{2} \left(65 {x}^{5} \sqrt{x} - \frac{21}{{x}^{4} \sqrt{x}}\right)$