# How do you differentiate v=(sqrtx+1/root3x)^2?

Apr 19, 2018

$v ' \left(x\right) = 1 + \frac{1}{3} {x}^{- \frac{5}{6}} - \frac{2}{3} {x}^{- \frac{5}{3}}$

#### Explanation:

write both $x$-terms as numerical exponents of $x :$

$\sqrt{x} = {x}^{\frac{1}{2}}$

$\sqrt[3]{x} = {x}^{\frac{1}{3}}$

$\frac{1}{\sqrt[3]{x}} = {x}^{- \frac{1}{3}}$

${\left(\sqrt{x} + \frac{1}{\sqrt[3]{x}}\right)}^{2} = {\left({x}^{\frac{1}{2}} + {x}^{- \frac{1}{3}}\right)}^{2}$

using FOIL:

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

then you can differentiate these values separately, using the power rule

$f \left(x\right) = n {x}^{n - 1}$.

when ${x}^{n} = {x}^{1}$, $n {x}^{n - 1}$ is ${x}^{0}$, or $1$.

when ${x}^{n} = {x}^{\frac{1}{6}}$, $n {x}^{n - 1}$ is $\frac{1}{6} {x}^{- \frac{5}{6}}$.

when ${x}^{n} = {x}^{- \frac{2}{3}}$, $n {x}^{n - 1}$ is $- \frac{2}{3} {x}^{- \frac{5}{3}}$.

hence, when ${x}^{1} + {x}^{\frac{1}{6}} + {x}^{\frac{1}{6}} + {x}^{- \frac{2}{3}}$ is differentiated, the result is

$1 + \frac{1}{6} {x}^{- \frac{5}{6}} + \frac{1}{6} {x}^{- \frac{5}{6}} + - \frac{2}{3} {x}^{- \frac{5}{3}}$.

this can be simplified to $1 + \frac{1}{3} {x}^{- \frac{5}{6}} - \frac{2}{3} {x}^{- \frac{5}{3}}$.

or even $\frac{3 + {x}^{- \frac{5}{6}} - 2 {x}^{- \frac{5}{3}}}{3}$.

$v ' \left(x\right) = 1 + \frac{1}{3} {x}^{- \frac{5}{6}} - \frac{2}{3} {x}^{- \frac{5}{3}}$.