# Derivatives

• Dec 5th 2008, 01:50 PM
McDiesel
Derivatives

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

I have no clue, especially with the nth root
• Dec 5th 2008, 01:58 PM
Chop Suey
Using the quotient rule:
$\left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$

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

I'll leave the simplification to you
• Dec 6th 2008, 07:48 AM
Simplifying indices
Hi -
Quote:

http://www.mathhelpforum.com/math-he...ec315d30-1.gif

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):

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

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

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

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

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

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

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

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

Hope that helps.