Calculus 15: Derivatives of Trigonometric Functions

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
$$\frac{d}{dx}\sin x=\cos x$$
In a similar manner, we can also obtain
$$\frac{d}{dx}\cos x=-\sin x$$
Using the reciprocal rule (baby quotient rule) here along with the derivatives of $\sin x$ and $\cos x$, we find the derivatives of $y=\sec x$, $y=\csc x$ as
\begin{align*}
\frac{d}{dx}\sec x&=\sec x\tan x\\
\frac{d}{dx}\csc x&=-\csc x\cot x
\end{align*}
Finally, using the quotient rule here along with the derivatives of $\sin x$ and $\cos x$, we find the derivatives of $y=\tan x$, $y=\cot x$ as \begin{align*}\frac{d}{dx}\tan x&=\sec^2 x\\
\frac{d}{dx}\cot x&=-\csc^2 x\end{align*}