Differentiate and simplify. $m(x) = \frac{x}{\sqrt{4x-3}}$ My work so far is:
\begin{align}
m'(x) &= \frac{(1)(\sqrt{4x-3})-(x)(1/2)(4x-3)^{-1/2}(4)}{(\sqrt{4x-3})^2} \\
&= \frac{\sqrt{4x-3} - 2x(4x-3)^{1/2}}{4x-3}
\end{align}
and now I'm stuck on how to simplify further
 A: So you have arrived at
$$\frac{(4x-3)^{1/2}-2x(4x-3)^{-1/2}}{4x-3}.$$
(There is a typo in the second displayed line.) One might decide this is simple enough. But it is useful to multiply top and bottom by $(4x-3)^{1/2}$. We end up with
$$\frac{(4x-3)-2x}{(4x-3)^{3/2}},$$
and we can simplify the top a little more.
A: In general we have $\frac{\partial}{\partial x} \frac{f(x)}{g(x)}= \frac{\frac{\partial f(x)}{\partial x} \times g(x) - \frac{\partial g(x)}{\partial x} \times f(x)}{g^2(x)}$. So in the case of your question, set $f(x)= x$ and $g(x) = (4x - 3)^{\frac{1}{2}}$. And we have:
$\frac{\partial}{\partial x} (x) = 1$ and
$\frac{\partial}{\partial x} ((4x - 3)^{\frac{1}{2}}) = \frac{1}{2} \times 4  \times (4x - 3)^{- \frac{1}{2}}= 2 \times (4x - 3)^{- \frac{1}{2}}$. So we have:
$\frac{\partial}{\partial x} \frac{x}{(4x - 3)^{\frac{1}{2}}}= \frac{1 \times (4x - 3)^{\frac{1}{2}} - 2 \times (4x - 3)^{- \frac{1}{2}} \times x }{4x - 3} = \frac{2x-3}{(4x - 3)^{\frac{3}{2}}}$.
And we are done.
A: In the right hand side of your second line, you wrote

$(4x-3)^{1/2}$

Note the exponent here is incorrect: it should still be $(4x-3)^{-1/2}$.  Let's write it that way and try to proceed.
$$m'(x)=\frac{\sqrt{4x-3}-2x(4x-3)^{-1/2}}{4x-3}=\frac{\sqrt{4x-3}-\frac{2x}{\sqrt{4x-3}}}{4x-3}$$
Now, in the numerator, we can simplify via finding a common denominator:
$$\sqrt{4x-3}-\frac{2x}{\sqrt{4x-3}}=\frac{4x-3-2x}{\sqrt{4x-3}}=\frac{2x-3}{\sqrt{4x-3}}$$
Putting this back in the numerator, we get:
$$m'(x)=\frac{\frac{2x-3}{\sqrt{4x-3}}}{4x-3}=\frac{2x-3}{(4x-3)^{3/2}}$$
This is as simplified as it's likely expected to be.
A: One method for differentiating $m(x)$ is treating $m(x)=\frac{f(x)}{g(x)}$ with 
$f(x)=x$ and $g(x)=\sqrt{4x-3}$  and using the quotient rule for differentiation.
Then $m'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}=\frac{\sqrt{4x-3}-(x\frac{2}{\sqrt{4x-3}})}{4x-3}$.
Simplyfying we obtain the desired result $m'(x)=\frac{2x-3}{\sqrt{(4x-3)^{3}}}$.
A: Another way which makes life easier when you face products, quotients, powers : logarithmic differentiation $$m=\frac{x}{\sqrt{4x-3}}\implies \log(m)=\log(x)-\frac 12 \log(4x-3)$$ Differentiate $$\frac{m'} m=\frac 1x-\frac 12 \times \frac 4{4x-3}=\frac{2x-3}{x(4x-3)}$$ $$m'=m\times \frac{2x-3}{x(4x-3)}=\frac{x}{\sqrt{4x-3}}\times \frac{2x-3}{x(4x-3)}=\frac{2x-3}{(4x-3)^{3/2}}$$
