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?
 A: In fact, $$12\times 3^{2x-1} = 12 \times 3^{2x} \times 3^{-1} = 12\times 3^{2x} \times \frac {1}{3} = 4\times 3^{2x} $$
We have used the fact that $a^{b+c} = a^b \times a^c$ and that  $a^{-c} = \frac {1}{a^c} $. Hope it helps.  
A: Check your algebra. In your solution, $3^{2x-1}=3^{2x}3^{-1}=\frac13\cdot 3^{2x}$, and $\frac13\cdot 12=4$.
A: You're almost there ;-)$$12\cdot 3^{2x-1}\cdot\ln3=4\cdot3\cdot 3^{2x-1}\cdot\ln3=4\cdot 3^{2x}\cdot\ln3$$
A: 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 $y'$
A: $$y^\prime=6\cdot3^{2x−1}\cdot\ln{3}\cdot(2x-1)^\prime$$
$$y^\prime=6\cdot3^{2x−1}\cdot\ln{3}\cdot 2$$
$$y^\prime=12\cdot3^{2x−1}\cdot\ln{3}$$
Now you can drop the $-1$ from the exponent:
$$y^\prime=12\cdot\frac{3^{2x}}{3}\cdot\ln{3}$$
$$y^\prime=4\cdot 3^{2x}\cdot\ln{3}$$
Basically, your answer is correct, but you just needed further simplification.
Note: $x^{a+b}=x^a\cdot x^b$ and $x^{-1}=1/x$
