Can someone please help me with the steps on this particular problem

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

I have no clue, especially with the nth root

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- Dec 5th 2008, 12:50 PMMcDieselDerivatives
Can someone please help me with the steps on this particular problem

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

I have no clue, especially with the nth root - Dec 5th 2008, 12:58 PMChop Suey
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 - Dec 6th 2008, 06:48 AMGrandadSimplifying indices
Hi -

Quote:

Can someone please help me with the steps on this particular problem

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

I have no clue, especially with the nth root

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.

Grandad