# How do you find the derivative of 4sqrt(x-3) + x^5/(2x)?

May 20, 2015

The first step here would be to write the equation in a way that is easier to understand:

$f \left(x\right) = 4 \sqrt{x - 3} + {x}^{5} / \left(2 x\right) = 4 {\left(x - 3\right)}^{\frac{1}{2}} + {x}^{4} / 2$

now we solve for $f ' \left(x\right)$

we can use the chain rule in the first section. The chain rule states that:

$\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

Naming $u = \left(x - 3\right)$ and then using the chain rule, we get:

$\textcolor{red}{\frac{d}{\mathrm{dx}} 4 {\left(x - 3\right)}^{\frac{1}{2}} = 2 {\left(x - 3\right)}^{- \frac{1}{2}} \left(1\right) = 2 {\left(x - 3\right)}^{- \frac{1}{2}}}$

and the second section would be:

$\textcolor{b l u e}{\frac{d}{\mathrm{dx}} {x}^{4} / 2 = 2 {x}^{3}}$

now we also know that

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} 4 {\left(x - 3\right)}^{\frac{1}{2}} + \frac{d}{\mathrm{dx}} {x}^{4} / 2$

so we can now just drop in our two results:

$f ' \left(x\right) = 2 {\left(x - 3\right)}^{- \frac{1}{2}} + 2 {x}^{3}$