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
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
Hi -
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.
Grandad