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Find $\frac{dy}{dx}$ at $x = 3$ when $y=12\sqrt{x}$.

I changed y = $12\sqrt{x}$ to $12x^{1/2}$ which after$\frac{dy}{dx}$ is $6x^{-1/2}$ therefore $\frac{dy}{dx}$ at $x = 3$ is $6(3)^{-1/2}$. Is it correct or are there any more steps after this or can I leave like that?

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    Yes, you're correct. – 2017-01-09

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It is correct;

One can, however, rewrite $6\cdot 3^{-\frac12}$:

$$6\cdot 3^{-\frac12} = 2\cdot3^1\cdot3^{-\frac12} = 2\cdot3^{\frac12} = 2\sqrt{3}$$

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    Excuse me if I'm being silly but how can I tell when it can be rewritten that way? – 2017-01-09
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    @user405455 not silly. I saw we were multiplying 6 by a negative exponent of 3 and I also remembered that 6 has 3 as a factor so I factored $6 = 2\cdot3$. Then if you recall that property of exponents: $a^p\cdot a^q = a^{p+q}$ you get what I wrote. – 2017-01-09
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That's correct. However I suggest you don't convert it to exponential, because it's something you will get into the habit of doing and then bigger derivatives will make your equation look ugly.

Just remember if $f(x) = (1)\sqrt{x}; f'(x) = \frac{1}{(2)(1)\sqrt{x}} $

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    The rule I used to give my students was to give an answer in the same symbols as the question. Thus if there are fractional exponents in the question, the answer would be written with fractional exponents. But if there were radical signs, the answer should be given using radicals. Negative exponents should never appear in an answer since they may hide a possible division by $0$: $3a^{-1}$ is not 0 when $a=0$. It doesn't exist. – 2017-01-10
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    Thats true aswell – 2017-01-10
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Yes, it is that simple, there are no other steps you should be required to show (given that you have derived derivatives from first principles or your instructor requests so)

$$f(x) = 12\sqrt{x} \implies \frac{df(x)}{dx} = \frac{6}{\sqrt{x}} \implies f'(3) = \frac{6}{\sqrt{3}}$$