Show $f$ is differentiable at $x=4$ $f$ is continuous. $f$ is defined on $(0,1) \cup (1, \infty)$.
$$f(x) = \begin{cases} \frac{4\sqrt x}{(\ln x)^2} & \text{if } x\in (0,1) \cup (1,e^4),\\
\frac{e^2}{64} (ln x) (8-lnx) & \text{ if } x\in[e^4, \infty) 
\end{cases}$$
Need to show that $f$ is differentiable at $x = e^4$
I tried differentiating and subbing in $e^4$ into $x$
$f'-(x) = 1/(4e^2)$
$f'+(x) = 0$

i will try this. make a change of variable $$x = e^4(1+h).$$  we can now write 
$$\begin{align} \ln x &= \ln \left(e^4(1+h) \right) = 4 + \ln(1+h) = 4 + h + \cdots\\
(\ln x)^{-2} &=(4+h \cdots)^{-2} = 4^{-2} - 24^{-3}h+ \cdots=\frac 1{16}-\frac h{32}+\cdots\\
\sqrt x &= e^2(1 + \frac 12 h + \cdots)= e^2 + \frac{e^2h}2+\cdots\\
\frac{e^2}{64}(\ln x)(8-\ln x) &= \frac{e^2}{64}(4 + h + \cdots)(4 - h + \cdots) = \frac{e^2}{4} + \cdots \tag 1
\end{align}$$
collecting all these into $$\frac{4\sqrt x}{(\ln x)^2} = 4(e^2 +\frac{e^2h}2+\cdots )(\frac 1{16}-\frac h{32}+\cdots) = 4(\frac{e^2}{16} + \cdots)=\frac{e^2}{4} + \cdots \tag 2$$
you can see from $(1)$ and $(2)$ that $f$ is differentiable at $x = 4$ and its derivative is zero.
