In this note, we study derivatives of trigonometric functions $y=\sin x$, $y=\cos x$, $y=\sec x$, $y=\csc x$, $y=\tan x$, and $y=\cot x$.  First we calculate the derivative of $y=\sin x$. \begin{align*}\frac{d}{dx}\sin x&=\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}\\&=\lim_{h\to 0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\&=\lim_{h\to 0}\left[\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\right]\end{align*} Recall that $\lim_{h\to 0}\frac{\cos h -1}{h}=0$ and $\lim_{h\to 0}\frac{\sin h}{h}=1$. Hence we obtain \begin{equation}\label{eq:dsin}\frac{d}{dx}\sin x=\cos x\end{equation} In a similar manner we can also obtain \begin{equation}\label{eq:dcos}\frac{d}{dx}\cos x=-\sin x\end{equation} Using the reciprocal rule (baby quotient rule) along with \eqref{eq:dsin} and \eqref{eq:dcos}, we find the derivatives of $y=\sec x$, $y=\csc x$ as \begin{align}\label{eq:d\sec}\frac{d}{dx}\sec x&=\sec x\tan x\\\label{eq:dcsc}\frac{d}{dx}\csc x&=-\csc x\cot x\end{align} Finally using the quotient rule along with \eqref{eq:dsin} and \eqref{eq:dcos}, we find the derivatives of $y=\tan x$, $y=\cot x$ as \begin{align}\label{eq:d\tan}\frac{d}{dx}\tan x&=\sec^2 x\\\label{eq:dcot}\frac{d}{dx}\cot x&=-\csc^2 x\end{align}