# Derivative of csc x: Formula, Proof and Examples

> The derivative of csc x is -csc x cot x. See the proof from 1/sin x, the chain rule for csc(u), a graph and worked examples like csc(3x).

Source: https://practicecalculus.com/derivatives/csc-x
Updated: 2026-11-06

$$ \frac{d}{dx}\,\csc x = -\csc x\cot x $$

## At a glance

- **Derivative:** $-\csc x\cot x$ for $x \ne k\pi$
- **In sin and cos:** $-\dfrac{\cos x}{\sin^2 x}$
- **Chain rule form:** $\dfrac{d}{dx}\csc u = -\csc u\cot u \cdot u'$
- **Slope at x = π/2:** $-\csc\frac{\pi}{2}\cot\frac{\pi}{2} = 0$
- **Integral version:** $\int \csc x\cot x\,dx = -\csc x + C$

## Key takeaways

- The derivative of $\csc x$ is $-\csc x\cot x$.
- It mirrors the derivative of $\sec x$: swap each function for its "co" partner and add a minus sign.
- With the chain rule, $\frac{d}{dx}\csc u = -\csc u\cot u\cdot u'$.
- Reading it backward gives $\int\csc x\cot x\,dx = -\csc x + C$.

The **derivative of $\csc x$ is $-\csc x\cot x$**. Cosecant is $\frac{1}{\sin x}$, and like every "co" trig function, its derivative carries a minus sign.

$$\frac{d}{dx}\,\csc x = -\csc x\cot x$$

## Graph of csc x and its derivative

```graph
title: 'Graph of y = csc x and its derivative y = -csc x cot x'
caption: 'One branch of $\csc x$ on $(0, \pi)$. It falls to a minimum at $x = \frac{\pi}{2}$, where the slope is $0$, then rises.'
x: [-0.3, 3.5]
y: [-6, 7]
functions:
  - expr: csc(x)
    label: y = csc x
    domain: [0.16, 2.98]
  - expr: -csc(x)cot(x)
    label: y = -csc x cot x (derivative)
    dashed: true
    domain: [0.38, 2.76]
vertical:
  - x: 3.1416
    label: x = π
points:
  - [1.5708, 1, 'minimum, slope 0']
```

## Proof with the chain rule

Write $\csc x = (\sin x)^{-1}$. With the [chain rule](/rules/chain-rule):

$$\frac{d}{dx}(\sin x)^{-1} = -(\sin x)^{-2}\cdot\cos x = -\frac{\cos x}{\sin^2 x}$$

Split the fraction:

$$-\frac{\cos x}{\sin^2 x} = -\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} = -\csc x\cot x$$

This is a special case of the [derivative of 1/x](/derivatives/1-over-x) pattern, $\frac{d}{dx}\frac{1}{u} = -\frac{u'}{u^2}$, with $u = \sin x$.

## A pattern worth knowing

The six trig derivatives come in pairs. Swap each function for its "co" partner and flip the sign:

| Function | Derivative | Partner | Derivative |
| --- | --- | --- | --- |
| $\sin x$ | $\cos x$ | $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ | $\cot x$ | $-\csc^2 x$ |
| $\sec x$ | $\sec x\tan x$ | $\csc x$ | $-\csc x\cot x$ |

So if you remember the [derivative of sec x](/derivatives/sec-x), you get the one for $\csc x$ for free.

## The chain rule version: derivative of csc(u)

$$\frac{d}{dx}\,\csc u = -\csc u\cot u\cdot u'$$

**Example 1.** $\frac{d}{dx}\csc(3x) = -3\csc(3x)\cot(3x)$.

**Example 2.** $\frac{d}{dx}\csc(x^2) = -2x\csc(x^2)\cot(x^2)$.

**Example 3.** $\frac{d}{dx}\csc\left(\frac{x}{2}\right) = -\frac{1}{2}\csc\frac{x}{2}\cot\frac{x}{2}$.

## More worked examples

**Example 4: a product.** $\frac{d}{dx}(x^2\csc x) = 2x\csc x - x^2\csc x\cot x$.

**Example 5: a power.** $\frac{d}{dx}\csc^2 x = 2\csc x\cdot(-\csc x\cot x) = -2\csc^2 x\cot x$.

**Example 6: a value.** At $x = \frac{\pi}{6}$, $\csc\frac{\pi}{6} = 2$ and $\cot\frac{\pi}{6} = \sqrt{3}$, so the slope is $-2\sqrt{3} \approx -3.46$. The curve is falling steeply there, which matches the graph.

## How to read the graph

Each branch of cosecant is a U shape sitting above or below the axis. On the upper branch between $0$ and $\pi$, the curve falls on the left half and rises on the right half, with its lowest point at $x = \frac{\pi}{2}$. The derivative follows the same story: negative on the left, zero at the bottom, positive on the right. Near the asymptotes at $0$ and $\pi$ the curve is almost vertical, so the derivative grows without bound in size there. Keeping this picture in mind is a fast way to check the sign of any answer you compute.

## Common mistakes

1. **Dropping the minus sign.** The "co" rule: $\cos$, $\cot$ and $\csc$ all have negative derivatives.
2. **Writing $-\cot^2 x$ or $-\csc^2 x$.** $-\csc^2 x$ is the derivative of $\cot x$. The derivative of $\csc x$ is $-\csc x\cot x$.
3. **Mixing up the reciprocal.** $\csc x = \frac{1}{\sin x}$, not $\frac{1}{\cos x}$.

The integral side is on the [integral of csc x](/integrals/csc-x) page. For more practice, try the [trig derivative problems](/practice-problems/derivatives-of-trig-functions).

## Frequently asked questions

### What is the derivative of csc x?

The derivative of $\csc x$ is $-\csc x\cot x$.

### How do you prove the derivative of csc x?

Write $\csc x = (\sin x)^{-1}$. The chain rule gives $-(\sin x)^{-2}\cos x = -\frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} = -\csc x\cot x$.

### What is the derivative of csc(2x)?

By the chain rule, $-2\csc(2x)\cot(2x)$.

### Is cosec x the same as csc x?

Yes. Both mean $\frac{1}{\sin x}$, and the derivative of cosec x is $-\operatorname{cosec}x\cot x$.