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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.

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Can you use proper formatting. – Priyatham Mar 9 '14 at 13:44
Your way of writing is extremely unclear and hard to read. At least use parentheses! Anyway, try to type mathematics with LaTeX-MathJax in this site. – DonAntonio Mar 9 '14 at 13:44
Hi. Sorry, I know. I'm typing it now, wait. : (9x^2 * 1/x) - (3x^3 * -x^-2), all this is the numerator, and then we have (1/(x^2)), which is the denominator. – Donny Mar 9 '14 at 13:49
Why would you use the quotient rule for $3x^3 / (1/x) = 3x^4$? – TMM Mar 9 '14 at 13:53
Because it's not about the derivative, it's about the simplifying method. I picked an easy derivative so I could check if I had simplified it correctly. Or have I misunderstood something and I can't use the quotient rule here at all? – Donny Mar 9 '14 at 14:00

Why won't you first simplify the $\;$ &@#^* $\;$ function??

$$f(x)=\frac{3x^3}{\frac1x}=3x^4\ldots !$$

And now just simply differentiate...

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That's the way. – Tom Collinge Mar 9 '14 at 14:48
I think the reason OP did not simplify the expression beforehand is that she wanted to construct an expression which would give her practice in simplifying the result, and which she could easily check to see if she had done it correctly. If so, then this is an excellent choice. – MJD Mar 9 '14 at 20:26
@MJD, too much guessing imo...and how do you know the OP is a "she" ? – DonAntonio Mar 9 '14 at 20:49
I misread the original question, and the remark that I thought OP had made isn't there, so I withdraw my earlier comment. – MJD Mar 9 '14 at 20:53

$$\dfrac{9x^2 \cdot \frac 1x - 3x^3 \cdot -x^{-2}}{1/x^2} = \dfrac{9x^2 \cdot \frac 1x - 3x^3 \cdot \frac{-1}{x^2}}{1/x^2} = \dfrac{9x +3x}{1/x^2} = x^2(12x) = 12x^3$$

Note, however that $$f(x) = \dfrac {3x^3}{\frac 1x} = x(3x^3) = 3x^4 \implies f'(x) = 12x^3$$

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@Mark I started my answer before the denominator was clarified. Now edited accordingly. – amWhy Mar 9 '14 at 14:04
You should add that $f'(x) = 12x^3$, only when $x\neq 0$, in which case it is undefined. – Sawarnik Mar 9 '14 at 15:57
@Sawarnik, you perhaps meant " The OP should add..." . It seems wise to always leave a little to complete for the OP. – DonAntonio Mar 9 '14 at 15:58

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, $x \neq 0$. $$\boxed{\therefore f^\prime(x)=12x^3, \ x \neq 0}$$

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