1. ## Derivatives

$\displaystyle \frac{3x^2-9x-5}{\sqrt[4]{x}}$

I have no clue, especially with the nth root

2. Using the quotient rule:
$\displaystyle \left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$

$\displaystyle \left(\frac{3x^2-9x-5}{x^\frac{1}{4}}\right)' = \frac{(6x-9)(x^\frac{1}{4})-(3x^2-9x-5)(x^{-\frac{3}{4}})}{x^\frac{2}{4}}$

I'll leave the simplification to you

3. ## Simplifying indices

Hi -

I have no clue, especially with the nth root
Begin by dividing each of the three terms in the numerator (top) by the term in the denominator.

Before you do this, here are a few things you'll need to know about handling indices (powers):

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

$\displaystyle x^{-a}=\frac{1}{x^a}$

$\displaystyle x^{a+b}=x^a\times x^b$

$\displaystyle x^{a-b}=x^a\div x^b$

First, then, divide $\displaystyle 3x^2$ by $\displaystyle \sqrt[4]{x}$ like this:

$\displaystyle 3x^2 \div \sqrt[4]{x} = 3x^2 \div x^\frac{1}{4}=3x^{(2-\frac{1}{4})}=3x^\frac{7}{4}$

Now divide $\displaystyle -9x$ by $\displaystyle \sqrt[4]{x}$, and then divide $\displaystyle -5$ by $\displaystyle \sqrt[4]{x}$ in the same way.

You'll then have three separate terms in $\displaystyle x$ which you can differentiate in the usual way.

Hope that helps.