New to derivation can someone explain the rule for finding the $f'(x)$ given this specific $f(x)$ function I am new to calculus in all honesty and we are looking at derivatives so $f'(x)$ given an $f(x)$ function. I understand the majority of the rules, but in the practise worksheet we were given, one the questions contained fractions times a power of $x$. I have no idea about any rule in order to derivative fractional functions.
The question concerned is as follows:
$$f(x) = \frac{2}{3} x^{3} + \frac{3}{2} x^{2} -2x + 4$$
what is the rule in abstract terms, so instead of the above what if i wrote it like this:
$$f(x) = \frac{n}{k} x^{p} + \frac{n}{k} x^{p} -2x + 4$$
what happens to the $n$, $k$ and $p$? im not sure on the correct terminology but what i mean is explain it i a way such as this:
if: $f(x) = x^n$ then $f'(x) = nx^ {(n-1)}$
Hopefully you understand what i am asking.
Also, i cant seem to get my equations to format properly :/ sorry about this.
Cheers,
Chris. 
 A: Here is my attempt at the answer based on people's suggestions can somebody tell me if i'm correct:
first of all we start with:
$$ f = \frac{2}{3} x^{3} + \frac{3}{2} x^{2} -2x + 4 \\\\ $$
which is the same as:
$$f = \frac{2(x^{3})}{3}  + \frac{3(x^{2})}{2} -2x + 4 \\\\$$
then simplify the equation (power rule):
$$\frac{2(x^{3})}{3} = \frac{2(3x^{2})}{3}  = 2x^{2} \\\\$$
$$
\frac{3(x^{2})}{2} = \frac{3(2x)}{2} = 3x \\\\$$
which now makes the function:
$$f(x)= 2x^{2} + 3x -2x + 4 \\\\$$
thus has a derivative of:
$$f'(x) = 2x^{2} + 3x -2 \\\\$$
Does anybody disagree with this?
A: Hint: If $f$ is a function and $c$ is a constant, then $g(x):=c\cdot f(x)$ is a function that is differentiable everywhere that $f$ is differentiable (possibly in more places, if $c=0$). Indeed, for any $h\ne 0,$ we have $$\frac{g(x+h)-g(x)}{h}=\frac{c\cdot f(x+h)-c\cdot f(x)}{h}=c\cdot\frac{f(x+h)-f(x)}{h}.$$ From this, it follows that if $f'(x)$ is defined, then so is $g'(x),$ and in particular, $g'(x)=c\cdot f'(x)$.
Another observation to make is that $(f+g)'(x)=f'(x)+g'(x),$ which we can show with similar manipulations of the difference quotient. Putting it all together with the power rule $(x^n)'=nx^{n-1},$ we can readily find the derivative of any polynomial function, regardless of whether the coefficients are integers, rational numbers, or otherwise.
A: The Power Rule is
$$y=x^n\implies \frac{d}{dx}=\left(nx^{n-1}\right)$$
The Constant Factor Rule states that if a function is being multiplied by a constant, then that constant does not affect its derivative. Meaning you can take the constant out and differentiate as normal and then multiply the derivative by the constant.
$$g(x)=kf(x) \implies g'(x)=kf'(x)$$
Okay let's take the term in $x^3$ and apply both rules.
$$y=\frac{2}{3}x^3$$
Using the Power Rule
$$\frac{d}{dx}=\left(\frac{6x^{3-1}}{3}\right)=2x^2$$
Using the Constant Factor Rule
$$\frac{d}{dx}=\frac{2}{3}\cdot \left(3x^{3-1}\right)=2x^2$$
The results are the same, so you can decide to use whichever feels more comfortable but for more complicated function it might make it easier if you took out the constant.
Answering your question, you can extend the power to incorporate fractions.
$$y=\frac{a}{b}x^n \implies \frac{d}{dx}=\frac{an}{b}x^{n-1}$$
