Lagrange Theorem with $\mathbb{tan}(x)$ I take $\mathbb{tan}(x)$ in $[30°, 45°]$ and I want to find $f'(c)$. The hypothesis are satisfied. I compute:
$$f'(c)=\frac{\mathbb{tan}(45°)-\mathbb{tan}(30°)}{45°-30°}=\frac{1-0.577..}{15°}\simeq0.0282$$
Now I know the derivate of $\mathbb{tan}(x)$ that is equal to $1+\mathbb{tan}^2(x)$, but if I try to calculate the point $c$ I obtain:
$$1+\mathbb{tan}^2(x)=0.0282$$ that is impossible.
If I use radians the result changes and it seems I can cumpute the point $c$:
$$f'(c)=\frac{1-0.577..}{\pi/12}\simeq1.6157$$
Someone could explain why I can compute with the radians but not with degrees?
 A: That's because the derivative of $\tan(x)$ is only $1 + \tan^2(x)$ when $x$ is expressed in radians. Let $r(x)$ be a function which inputs $x$ in degrees and outputs $r(x)$ in radians. So $r(30^\circ) = \pi/6$, for example. The general formula is $r(x) = \frac{\pi}{180}x$. So using the chain rule, 
$$ \tan'(r(x)) = (1 + \tan^2(r(x))) r'(x) = \frac{\pi}{180} (1 + \tan^2(r(x))).$$
So, you're missing the extra factor of $\pi/180$ in the derivative.
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The reason why this is true goes back to the definition of the derivative. Take $\sin x$. Calculating its derivative at $x = 0$ with the difference quotient formula gives
$$ \lim_{h \to 0} \frac{\sin h - \sin 0}{h} = \lim_{h \to 0} \frac{\sin h}{h} = 1 .$$
The way this limit is computed is geometrically, it'll be on the internet somewhere. The takeaway is this limit is only equal to $1$ when $h$ is measured in radians. That is why we pick radians as the "natural" way to measure angles in math. If you pick degrees instead, there will be an extra factor of $\frac{\pi}{180}$ floating around, as we saw above. 
A: A possible way to understand what is going on is to think a little about what the quantities represent.  One nice description of an angle is that it corresponds to an arc length on the unit circle.  That is, the measure of an angle $\theta$ is exactly the length of the arc on the unit circle subtended by $\theta$ assuming that $\theta$ is measured in radians.  The sine and cosine can be interpreted as signed lengths (of the legs of the triangle with hypotenuse along $\theta$), and the remaining trigonometric functions can be written and analyzed in terms of sines and cosines.
The next step is to compute the derivative.  By definition
\begin{align} \frac{\mathrm{d}}{\mathrm{d}\theta} \sin(\theta)
&= \lim_{h\to 0} \frac{\sin(\theta + h) - \sin(\theta)}{h} \\
&= \lim_{h\to 0} \frac{\sin(\theta)\cos(h) + \cos(\theta)\sin(h) - \sin(\theta)}{h} && (\text{angle addition}) \\
&= \sin(\theta) \lim_{h\to 0} \frac{\cos(h)-1}{h} + \cos(\theta)\lim_{h\to 0} \frac{\sin(h)}{h}. \tag{1}
\end{align}
Here is where things get a little tricky.  We need to compute
$$\lim_{h\to 0} \frac{\sin(h)}{h}. $$
A standard trick is to apply the squeeze theorem.  Consider the following figure:

If $h$ is measured in radians, then the measure of the angle $h$ is the length of the arc subtended by $h$, i.e. the arc $\overset\frown{DA}$.  This is really the most important point, so let me state it again:

The length of the arc $\overset{\frown}{DA}$ (in units of length) is equal to the measure of the angle $h$ (measured in radians).

This will not be true if we measure $h$ in degrees.  If we measure $h$ in degrees, then the length of the arc will be $\frac{h}{360} \cdot 2\pi$, where $\frac{h}{360}$ is the fraction of the circle subtended by $h$, and $2\pi$ is the circumference of the circle.
It is true (though perhaps not completely obvious and trivial) that $\overline{AE}$ is shorter than the arc.  But the length of that leg is, by definition, $\sin(h)$, so we have that
$$ \sin(h) \le h. $$
On the other hand, the segment $\overline{DF}$ is longer than the arc.  By properties of similar triangles, we have
$$ \frac{FD}{AE} = \frac{BD}{BE}
\implies FD = AE \frac{BD}{BE} = \sin(h) \frac{1}{\cos(h)} = \frac{\sin(h)}{\cos(h)}. $$
Therefore
$$h \le \frac{\sin(h)}{\cos(h)}. $$
Combining this with the previous estimate and taking limits (using the squeeze theorem), we get that
$$ \sin(h) \le h \le \frac{\sin(h)}{\cos(h)}
\implies 1 \le \frac{h}{\sin(h)} \le \cos(h)
\implies 1 \le \lim_{h\to 0} \frac{h}{\sin(h)} \le \lim_{h\to 0} \cos(h) = 1.$$
This tells use that
$$ \lim_{h\to 0} \frac{h}{\sin(h)} = 1 \implies \lim_{h\to 0} \frac{\sin(h)}{h} = 1. $$
From this, an application of the Pythagorean identity allows us to conclude that
$$ \lim_{h\to 0} \frac{\cos(h)-1}{h} = 0. $$
Substituting these back into (1), we finally obtain
\begin{align} \frac{\mathrm{d}}{\mathrm{d}\theta} \sin(\theta)
&= \sin(\theta) \lim_{h\to 0} \frac{\cos(h)-1}{h} + \cos(\theta)\lim_{h\to 0} \frac{\sin(h)}{h} \\
&= \sin(\theta) \cdot 0 + \cos(\theta) \cdot 1 \\
&= \cos(\theta).
\end{align}
The key point here is that this argument only hangs together if we consider how the length of an arc on the unit circle corresponds to the measure of the subtending angle.  When we measure the angle in radians, the arc and the angle have the same measure, and we get the result we expect, i.e.
$$ \frac{\mathrm{d}}{\mathrm{d}\theta} \sin(\theta) = \cos(\theta). $$
From this identity, we can deduce the remaining trigonometric derivatives.  If the angle is not measured in radians, then the first thing to do is make the change of variables to that it is measured in radians.

So, what does this have to do with your problem?  The derivatives of all of the trigonometric functions are derived with angles measured in radians.  Thus if you have an angle $\theta$ measured in degrees, the first thing that you have to do to work with it is convert to radians.  Define the function
$$ u(\theta) = \frac{\pi}{180} \theta. $$
If $\varphi$ is the degree measure of an angle, then $u(\varphi)$ is the radian measure of the same angle.  It then follows from the chain rule that
$$ \frac{\mathrm{d}}{\mathrm{d}\theta} \tan(u(\theta))
= (1+\tan(u(\theta))^2) u'(\theta)
= \left(1+\tan\left( \frac{\pi}{180}\theta \right)^2\right) \frac{\pi}{180}. $$
Note that you end up with two factors of $\frac{\pi}{180}$:  one in the argument of the trigonometric function, and on as an extra scaling at the end.
