Q:

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.

