Simplify this, how to? The original function is $$f(x)=\cfrac{3x^3}{\cfrac{1}{x}} $$
It's the quotient rule I'm using so: 
$$ f^\prime\left(x\right) =\dfrac{9x^2\cdot\cfrac{1}{x} -  3x^3 \cdot \left(-x^{-2}\right)}{\dfrac{1}{x^2}}$$ 
I suck at simplifying, so I just need to get the hang of it. I've done some already, but this one is troubling me. Thanks. Oh, and STEP BY STEP please. 
 A: $$\dfrac{9x^2 \cdot \frac 1x - 3x^3 \cdot -x^{-2}}{1/x^2} = \dfrac{9x^2 \cdot \frac 1x - 3x^3 \cdot \frac{-1}{x^2}}{1/x^2} = \dfrac{9x +3x}{1/x^2} = x^2(12x) = 12x^3$$
Note, however that $$f(x) = \dfrac {3x^3}{\frac 1x} = x(3x^3) = 3x^4 \implies f'(x) = 12x^3$$
A: Our original equation:
$$f(x)=\dfrac{3x^3}{\left(\dfrac{1}{x}\right)}$$
We need to find $f^\prime(x)$.
Step 1: Define the restriction(s). The derivative is invalid at this/these point(s).
The restriction is:
$$x \neq 0$$
because division by $0$ is not allowed.
Step 2: Simplify $f(x)$. We will use the fact that $\dfrac{a}{c}=a\left(\dfrac{1}{c}\right)$.
$$f(x)=\dfrac{3x^3}{\left(\dfrac{1}{x}\right)}$$
$$f(x)=3x^3\left(\dfrac{x}{1}\right)$$
$$f(x)=3x^3(x)$$
$$f(x)=3x^4$$
Step 3: Use the power rule to find $f^\prime(x)$.
The power rule states that for any function $f(x)=x^n$, it's derivative is $f^\prime(x)=nx^{n-1}$.
$$f(x)=3x^4$$
$$\dfrac{d}{dx}f(x)=\dfrac{d}{dx}3x^4$$
$$\dfrac{d}{dx}3x^4=3\dfrac{d}{dx}x^4 \ \ \text{(Constant rule)}$$
$$3\dfrac{d}{dx}x^4=3(4x^3)$$
$$3(4x^3)=12x^3$$
Remember, $x \neq 0$.
$$\boxed{\therefore f^\prime(x)=12x^3, \ x \neq 0}$$
A: Why won't you first simplify the $\;$ &@#^* $\;$ function??
$$f(x)=\frac{3x^3}{\frac1x}=3x^4\ldots !$$
And now just simply differentiate...
