How to simplify this differentiated equation further? The question I have is sqrt(2x-1) / 2e^3x and I am asked to differentiate and simplify.
This is my working below:

I am unsure of how to get to simplify this to get the final answer in the textbook shown below:

What do I have to do?
 A: Starting from your result, since $\frac{e^{3x}}{3^{6x}} = e^{-3x}$, $(2x - 1)^{\frac{1}{2}} = (2x - 1)^{-\frac{1}{2}}(2x - 1)$ and $(2x - 1)^{-\frac{1}{2}} = \frac{1}{\sqrt{2x - 1}}$,
$$\begin{equation}\begin{aligned}
& \frac{2e^{3x}\left((2x - 1)^{-\frac{1}{2}} - 3(2x - 1)^{\frac{1}{2}}\right)}{4e^{6x}} \\
& = \frac{e^{-3x}\left((2x - 1)^{-\frac{1}{2}} - 3(2x - 1)^{-\frac{1}{2}}(2x - 1)\right)}{2} \\
& = \frac{e^{-3x}(2x - 1)^{-\frac{1}{2}}\left(1 - 3(2x - 1)\right)}{2} \\
& = \frac{e^{-3x}\left(1 - 6x + 3)\right)}{2\sqrt{2x - 1}} \\
& = \frac{e^{-3x}\left(2\left(2 - 3x\right)\right)}{2\sqrt{2x - 1}} \\
& = \frac{\left(2 -3x\right)e^{-3x}}{\sqrt{2x - 1}}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This matches the final result shown for $g'(x)$ in $242$ in your image.
A: A small trick !
When you face expressions which only contain products, quotients, powers, exponentials, etc., logarithmic differentiation makes life much easier.
For your problem
$$y=\frac {\sqrt{2x-1}} {2e^{3x}}\implies \log(y)=\frac 12 \log(2x-1)-\log(2)-3x$$ Differentiating both sides
$$\frac {y'} y=\frac 12 \frac 2 {2x-1}-3= \frac 1 {2x-1}-3=\frac {2(2-3x)} {2x-1}$$ Now
$$y'=y \times\frac {y'} y=\frac {\sqrt{2x-1}} {2e^{3x}}\times \frac {2(2-3x)} {2x-1}=\frac{2-3x}{e^{3x} \sqrt{2x-1}}$$
