Why is $(\sec x)' = \tan x\sec x$ and not $\tan x$? As far as I understood, the Fundamental Theorem of Calculus states that the integral of a function is its anti-derivative. And yet, although the integral of $\tan x$ is $\sec x$, the derivative of $\sec x$ is $\tan x\sec x$. 
I understand the calculation and you get $\tan x\sec x$ as the derivative, but how does it make sense in light of the fundamental theorem? What am I missing here?

The actual anti-derivative of $\tan{x}$ is:
$$\int\tan{x}\,\mathrm{d}x=\int\frac{\sin{x}}{\cos{x}}\,\mathrm{d}x=\int\frac{-\mathrm{d}(\cos{x})}{\cos{x}}=-\ln{(\cos{x})}+\text{constant}.$$
This gives us the definite integral,
$$\int_{0}^{x}\tan{u}\,\mathrm{d}u=-\ln{(\cos{x})}=\ln{\left(\frac{1}{\cos{x}}\right)}=\ln{(\sec{x})}.$$
Applying the fundamental theorem of calculus to this integral gives us:
$$\begin{align}
\frac{d}{dx}\int_{0}^{x}\tan{u}\,\mathrm{d}u&=\frac{d}{dx}\ln{(\sec{x})}\\
\implies \tan{x}&=\frac{\frac{d}{dx}(\sec{x})}{\sec{x}}.
\end{align}$$
Multiplying both sides by $\sec{x}$ yields:
$$\frac{d}{dx}(\sec{x})=\tan{x}\sec{x}.$$
