5.8 Derivatives of The Trigonometric Functions

Before we start, recall that $$\begin{array}{}\text{(a)}& \underset{x\to 0}{lim}\frac{\sin x}{x}=0,\end{array}$$ also that
$$\begin{array}{rl}\underset{x\to 0}{lim}\frac{\cos x-1}{x}& =\underset{x\to 0}{lim}\frac{2{\sin }^2\frac{x}{2}}{x}\\ & =\underset{x\to 0}{lim}(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\sin \frac{x}{2})\\ & =\underset{x\to 0}{lim}\frac{\sin \frac{x}{2}}{\frac{x}{2}}\times \underset{x\to 0}{lim}\sin \frac{x}{2}\\ \text{(}\text{let }u=\frac{x}{2}\text{)}& & =\underset{u\to 0}{lim}\frac{\sin u}{u}\times \underset{x\to 0}{lim}\sin u& =1\times 0\\ \text{(b)}& & =0,\end{array}$$
and
$$\begin{array}{rl}\sin (A+B)& =\sin A\cos B+\cos A\sin B\\ \cos (A+B)& =\cos A\cos B-\sin A\sin B.\end{array}$$

$$\begin{array}{rl}\frac{dy}{dx}& =\underset{h\to 0}{lim}\frac{\sin (x+h)-\sin x}{h}\\ & =\underset{h\to 0}{lim}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ & =\underset{h\to 0}{lim}(\sin x\frac{\cos h-1}{h})\underset{h\to 0}{lim}(\cos x\frac{\sin h}{h})\\ & =\sin x\times \underset{h\to 0}{lim}\frac{\cos h-1}{h}+\cos x\times \underset{h\to 0}{lim}\frac{\sin h}{h}\\ & =\sin x\times 0+\cos x\times 1\\ & =\cos x.\end{array}$$

Therefore

$$\begin{array}{rl}\frac{dy}{dx}& =\underset{h\to 0}{lim}\frac{\cos (x+h)-\cos x}{h}\\ & =\underset{h\to 0}{lim}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}\\ & =\underset{h\to 0}{lim}\frac{\cos x(\cos h-1)-\sin x\sin h}{h}\\ & =\left(\underset{h\to 0}{lim}\frac{\cos x(\cos h-1)}{h}\right)-\left(\underset{h\to 0}{lim}\frac{\sin x\sin h}{h}\right)\\ & =\cos x\left(\underset{h\to 0}{lim}\frac{\cos h-1}{h}\right)-\sin x\left(\underset{h\to 0}{lim}\frac{\sin h}{h}\right)\\ & =\cos x\times 0-\sin x\times 1\\ & =-\sin x\end{array}$$

Therefore

Thus

$$\begin{array}{rl}\frac{d}{dx}\tan x& =\frac{d}{dx}\frac{\sin x}{\cos x}\\ & =\frac{\cos x\cos x-(-\sin x)\sin x}{{\cos }^{2}x}\\ & =\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}\\ & =\frac{1}{{\cos }^{2}x}\\ & ={\sec }^{2}x.\end{array}$$
That is, $$\frac{d}{dx}\tan x={\sec }^{2}x.$$ Also we recall that
$$\frac{1}{{\cos }^{2}x}=\frac{{\sin }^2+{\cos }^{2}x}{{\cos }^{2}x}={\tan }^{2}x+1.$$
Therefore, we can also write

$$\begin{array}{rl}\frac{d}{dx}\cot x& =\frac{d}{dx}\frac{\cos x}{\sin x}\\ & =\frac{-\sin x\times \sin x-\cos x\times \cos x}{{\sin }^{2}x}\\ & =-\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x}\\ & =-\frac{1}{{\sin }^{2}x}\\ & =-{\csc }^{2}x.\end{array}$$

$$\frac{d}{dx}\cot x=-{\csc }^{2}x$$ Because $$\frac{1}{{\sin }^{2}x}=\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x}=1+{\cot }^{2}x,$$ we have

$$\begin{array}{rl}\frac{d}{dx}\sec x& =\frac{d}{dx}\frac{1}{\cos x}\\ & =\frac{0\times \cos x-(-\sin x)\times 1}{{\cos }^{2}x}\\ & =\frac{\sin x}{{\cos }^2}\\ & =\frac{\sin x}{\cos x}\frac{1}{\cos x}\\ & =\tan x\sec x\end{array}$$

$$\begin{array}{rl}\frac{d}{dx}\csc x& =\frac{d}{dx}\frac{1}{\sin x}\\ & =\frac{0\times \sin x-\cos x\times 1}{{\sin }^{2}x}\\ & =\frac{-\cos x}{{\sin }^{2}x}\\ & =-\frac{\cos x}{\sin x}\frac{1}{\sin x}\\ & =-\cot x\csc x.\end{array}$$