<s>[INST] Confusion concerning logarithmic differentiation The problem is stated as follows: 
Find the equation of the line perpendicular to $f(x)=\pi^2+2^x+x^2+x^{1/x}$ at $x=1$.
My approach was this:
Notice that $f(1)=4+\pi^2$ so I am interested in the point $P(1, 4+\pi^2)$
Then taking the derivative
\begin{align}
\frac{df}{dx}&=\frac{d}{dx}\pi^2+\frac{d}{dx}2^x+\frac{d}{dx}x^2+\frac{d}{dx}x^{1/x}\\
&=2^x\ln{2}+2x+\frac{d}{dx}e^{1/x\ln{x}}\\
&=2^x\ln{2}+2x+x^{1/x}\cdot\left(-\frac{1}{x^2}\ln{x}+\frac{1}{x^2}\right)\\
\end{align}
Now finding $f'(1)$:
\begin{align}
f'(1)&=2\ln{2}+2+1\cdot\left(-\frac{1}{1^2}\ln{1}+\frac{1}{1^2}\right)\\
&=2\ln{2}+3
\end{align}
Now taking the opposite reciprocal gives
\begin{align}
-\frac{1}{2\ln{2}+3}
\end{align}
Now using point $P$ we get the following equation
\begin{align}
y-(4+\pi^2)&=-\frac{1}{2\ln{2}+3}\cdot(x-1)\\
y&=-\frac{(x-1)}{2\ln{2}+3}+4+\pi^2
\end{align}
I am quite certain that this is the correct answer. However, when I try a different method, namely logarithmic differentiation I get a different derivative:
Taking the natural log of both sides:
\begin{align}
\ln{f(x)}=\ln\pi^2+\ln{2^x}+\ln{x^2}+\ln{x^{1/x}}\\
\end{align}
And differentiating both sides:
\begin{align}
\frac{f'(x)}{f(x)}&=\frac{d}{dx}\ln{\pi^2}+\frac{d}{dx}x\cdot\ln{2}+\frac{d}{dx}2\ln{x}+\frac{d}{dx}\frac{1}{x}\ln{x}\\
f'(x)&=\left(\ln{2}+\frac{2}{x}+\frac{1}{x^2}-\frac{1}{x^2}\ln{x}\right)\cdot(\pi^2+2^x+x^2+x^{1/x})
\end{align}
Clearly something has gone very wrong here. Is there a flaw in my original answer or am I misunderstanding the logarithmic differentiation? Any help is appreciated, thanks!
 [/INST] $$f(x)=\pi^2+2^x+x^2+x^{1/x}$$
$$\ln |f(x)|=\ln |\pi^2+2^x+x^2+x^{1/x}|$$
You don't have a product of factors on RHS.This line is not correct:
$$\ln{f(x)}\ne \ln\pi^2+\ln{2^x}+\ln{x^2}+\ln{x^{1/x}}\\$$
</s>