# How do you find f'(x) using the definition of a derivative y=6e^x+4/root3x?

Feb 23, 2016

6e^x-4/(3x^(4/3

#### Explanation:

$\setminus \frac{d}{\mathrm{dx}} \left(6 {e}^{x} + \setminus \frac{4}{\setminus \sqrt{3} \left\{x\right\}}\right)$

Applying sum/difference rule: ${\left(f \setminus \pm g\right)}^{'} = {f}^{'} \setminus \pm {g}^{'}$

$= \setminus \frac{d}{\mathrm{dx}} \left(6 {e}^{x}\right) + \setminus \frac{d}{\mathrm{dx}} \left(\setminus \frac{4}{\setminus \sqrt{3} \left\{x\right\}}\right)$

We have,
$= \setminus \frac{d}{\mathrm{dx}} \left(6 {e}^{x}\right)$ = $6 {e}^{x}$

(Taking the constant out : ${\left(a \setminus \cdot f\right)}^{'} = a \setminus \cdot {f}^{'}$ i.e $= 6 \setminus \frac{d}{\mathrm{dx}} \left({e}^{x}\right)$ and applying common derivative of $\setminus \frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$)

And,

$\setminus \frac{d}{\mathrm{dx}} \left(\setminus \frac{4}{\setminus \sqrt{3} \left\{x\right\}}\right) = - \setminus \frac{4}{3 {x}^{\setminus \frac{4}{3}}}$

(Taking the constant out : ${\left(a \setminus \cdot f\right)}^{'} = a \setminus \cdot {f}^{'}$ i.e $= 4 \setminus \frac{d}{\mathrm{dx}} \left(\setminus \frac{1}{\setminus \sqrt{3} \left\{x\right\}}\right)$ and applying power rule for $= 4 \setminus \frac{d}{\mathrm{dx}} \left({x}^{- \setminus \frac{1}{3}}\right)$ i.e $\setminus \frac{d}{\mathrm{dx}} \left({x}^{a}\right) = a \setminus \cdot {x}^{a - 1}$ we get $= - \setminus \frac{4}{3 {x}^{\setminus \frac{4}{3}}}$)

So, finally
$6 {e}^{x}$ - $\setminus \frac{4}{3 {x}^{\setminus \frac{4}{3}}}$