<s>[INST] Differentiate $y = 6 \cdot 3^{2x - 1}$ I'm trying to differentiate  $y = 6 \cdot 3^{2x - 1}$, not really sure about my answer. I tried using wolfram alpha, but it doesn't really help me... I feel I'm kind of close, but I just seem to not be able to make the final leap.
So: $\frac{d}{dx}(6 \cdot 3^{2x - 1}) = 6 \frac{d}{dx}(3^{2x - 1}) = 6 \cdot 3^{2x - 1} \ln (3) \cdot 2 = 12 \cdot 3^{2x - 1} \ln (3)$
Now, according to the book, the answer would be: $4 \cdot 3^{2x}\ln (3)$.
What seems obvious is just dividing by $3$, but then I get in to trouble trying to apply it to $\frac{3^{2x - 1}}{3} = 3^{2x - 1} * 3^{-1} = 3^{2x - 2}$...
So how to proceed now?
 [/INST] We can clear up the situation by starting from the get go, with the reminder that for real $a, b, c$, we have $a^b\cdot a^c = a^{b+c}$.
$$y = 6 \cdot 3^{2x - 1} = 2\cdot 3\cdot 3^{2x - 1} = 2\cdot {3^1\cdot 3^{2x - 1}} = 2\cdot 3^{1 + 2x - 1} = 2\cdot 3^{2x}$$
Now find $y'$
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