Derivatives of Trigonometric Functions

Function	Derivative
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	${\sec }^{2}x=1+{\tan }^{2}x$
$\cot x$	$-{\csc }^{2}x=-(1+{\cot }^{2}x)$
$\sec x$	$\sec x\tan x$
$\csc x$	$-\csc x\cot x$

Prerequisites

Before computing these derivatives, recall:

\lim_{x\to0}\frac{\sin x}{x}=1\tag{a}\lim_{x\to0}\frac{\cos x-1}{x}=0\tag{b}

Proof of (b):

\begin{aligned} \lim_{x\to0}\frac{\cos x-1}{x}&=\lim_{x\to0}\frac{2\sin^2(x/2)}{x}=\lim_{x\to0}\frac{\sin(x/2)}{x/2}\cdot\sin\frac{x}{2}=1\times0=0 \end{aligned}

Also recall the angle-addition formulas:

$$\sin (A+B)=\sin A\cos B+\cos A\sin B$$$$\cos (A+B)=\cos A\cos B-\sin A\sin B$$

Derivative of $y=\sin x$

\begin{aligned} \frac{dy}{dx}&=\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}\\ &=\lim_{h\to0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ &=\sin x\cdot\lim_{h\to0}\frac{\cos h-1}{h}+\cos x\cdot\lim_{h\to0}\frac{\sin h}{h}\\ &=\sin x\cdot0+\cos x\cdot1=\cos x \end{aligned}\frac{d}{dx}\sin x=\cos x\tag{c}

Derivative of $y=\cos x$

Method (a): limit definition.

\begin{aligned} \frac{dy}{dx}&=\lim_{h\to0}\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}\\ &=\cos x\cdot\lim_{h\to0}\frac{\cos h-1}{h}-\sin x\cdot\lim_{h\to0}\frac{\sin h}{h}\\ &=\cos x\cdot0-\sin x\cdot1=-\sin x \end{aligned}

Method (b): using the Chain Rule and $\cos x=\sin (\pi /2-x)$:

$$\frac{d}{dx}\cos x=(\cos u)(-1){|}_{u=\pi /2-x}=-\cos (\pi /2-x)=-\sin x$$\frac{d}{dx}\cos x=-\sin x\tag{d}

• Both $\sin x$ and $\cos x$ are differentiable everywhere.
• You must memorize the derivatives of sine and cosine.

Derivative of $y=\tan x$

Write $\tan x={\displaystyle \frac{\sin x}{\cos x}}$ and apply the Quotient Rule:

\begin{aligned} \frac{d}{dx}\tan x&=\frac{\cos x\cdot\cos x-(-\sin x)\cdot\sin x}{\cos^2 x}=\frac{\cos^2 x+\sin^2 x}{\cos^2 x}=\frac{1}{\cos^2 x}=\sec^2 x \end{aligned}

Because $1/{\cos }^{2}x=({\sin }^{2}x+{\cos }^{2}x)/{\cos }^{2}x={\tan }^{2}x+1$:

$$\frac{d}{dx}\tan x={\sec }^{2}x=1+{\tan }^{2}x$$

Derivative of $y=\cot x$

\begin{aligned} \frac{d}{dx}\cot x&=\frac{d}{dx}\frac{\cos x}{\sin x}=\frac{-\sin x\cdot\sin x-\cos x\cdot\cos x}{\sin^2 x}=-\frac{1}{\sin^2 x}=-\csc^2 x \end{aligned}

Because $1/{\sin }^{2}x=({\sin }^{2}x+{\cos }^{2}x)/{\sin }^{2}x=1+{\cot }^{2}x$:

$$\frac{d}{dx}\cot x=-{\csc }^{2}x=-(1+{\cot }^{2}x)$$

Derivative of $y=\sec x$

\begin{aligned} \frac{d}{dx}\sec x&=\frac{d}{dx}\frac{1}{\cos x}=\frac{0\cdot\cos x-(-\sin x)\cdot1}{\cos^2 x}=\frac{\sin x}{\cos^2 x}=\tan x\sec x \end{aligned}$$\frac{d}{dx}\sec x=\tan x\sec x$$

Derivative of $y=\csc x$

\begin{aligned} \frac{d}{dx}\csc x&=\frac{d}{dx}\frac{1}{\sin x}=\frac{-\cos x}{\sin^2 x}=-\cot x\csc x \end{aligned}$$\frac{d}{dx}\csc x=-\cot x\csc x$$

Worked Examples

Frequently Asked Questions

Why does the derivative of cosine have a minus sign?

The cosine function is decreasing on $(0,\pi )$ where sine is positive, so its derivative must be negative on that interval. This matches ${\displaystyle \frac{d}{dx}}\cos x=-\sin x$. Geometrically, the tangent line to $y=\cos x$ has negative slope when .

Do these derivatives require $x$ to be in radians?

Yes. The limit $\underset{x\to 0}{lim}(\sin x)/x=1$ holds only when $x$ is in radians. If $x$ is in degrees, extra conversion factors appear, making the formulas much messier.

How can I remember whether (\sec x)' involves $\tan x$ or $\cot x$?

Remember the pairing: secant goes with tangent (both involve cosine in the denominator), and cosecant goes with cotangent (both involve sine). Both derivatives also include a copy of the original function: (\sec x)'=\sec x\tan x and (\csc x)'=-\csc x\cot x.