Using first principle method to get derivative of $\sin(x°)$ I saw the derivation of the derivative of $\sin x$ when $x$ was in radians but I don't know why we can't use the same derivation to get the derivative of $\sin(x°)$.
Here's my attempt:
Note that $x$ and $h$ used below are in degrees.
Let $f(x)=\sin(x)$. So
$$f'(x)= \lim_{h\rightarrow 0}\frac{\sin(x+h) - \sin x}{h}$$
Now,
$$f'(x) = \lim_{h\rightarrow 0} \frac{2\cdot \cos(x+\frac{h}{2})\cdot \sin(\frac{h}{2})}{h}$$
$\Rightarrow$ $f'(x) = \cos(x)\cdot \lim_{h\rightarrow 0}\frac{\sin(\frac{h}{2})}{(\frac{h}{2})}$
Since $h$ is also in degree we get
$f'(x)= \cos (x)$ where $x$ is in degree.
Where am I wrong ?
 A: The problem arises because of the imprecise language used to describe angle measurements. The expression "$\sin(x)$, where $x$ is measured in degrees" makes little sense when you take a step back and think about it. As Michael Spivak puts it in Calculus, "a number $x$ is simply a number—it does not carry a banner indicating that it is 'in degrees' or 'in radians'." The definition of $\sin(x)$ is simple: it is the vertical  coordinate that you arrive at after tracing an arc of $x$ units counterclockwise around the unit circle, starting from the position $(0,1)$. What people call "$\sin(x)$, where $x$ is measured in degrees" is actually a different function that is sometimes written as $\sin°$. By definition, $\sin°(x)=\sin\left(\frac{2\pi}{360}x\right)$. If we agree that the symbol $°$ means $\frac{2\pi}{360}$, then $\sin°(x)=\sin(x°)$. It is easy to compute $(\sin°)'$ using the chain rule, but it is also instructive to use differentiation from first principles:
\begin{align}
(\sin°)'(x) &= \cos(x)\cdot \lim_{h \to 0}\frac{\sin°(h)}{h} \\[5pt]
&= \cos(x) \cdot \lim_{h \to 0}\frac{\sin\left(\frac{2\pi}{360}h\right)}{h} \\[5pt]
&= \cos(x) \cdot \lim_{h\to0}\frac{\sin(\frac{2\pi}{360}h)}{\frac{2\pi}{360}h} \cdot \frac{2\pi}{360} \\[5pt]
&= \frac{2\pi}{360} \cdot \cos(x) \, .
\end{align}
The mistake you made occurred when you said "$h$ is also measured in degrees"; to reiterate, $h$ is a real number, not a physical quantity like length or weight.
