Derivative of csc – Derivation, Explanation, and Example

Learning about the derivative of $\boldsymbol{\csc x}$ will help us differentiate functions that contain $\csc x$ or the cosecant of another function. Understanding the components of $\dfrac{d}{dx} \csc x$ will also be essential in integral calculus. Hence, we must master this rule at this point.

The derivative of $\boldsymbol{\csc x}$ returns the product of $\boldsymbol{\csc x}$ and $\boldsymbol{\cot x}$. We can prove this derivative by differentiating $\boldsymbol{\dfrac{1}{\sin x}}$.

In this article, we’ll learn the following concepts that involve $\csc x$ and its derivative:

  • Familiarize yourself with the formula for $\dfrac{d}{dx} \csc x$.
  • Understand how to apply the quotient rule to derive the formula for $\dfrac{d}{dx} \csc x$.
  • Learn how to integrate this rule with the other fundamental and trigonometric derivative rules we’ve learned in the past.

We’ll begin our discussion by understanding the components that make up $\dfrac{d}{dx} \csc x$.

What is the derivative of csc?    

The derivative of $\csc x$ has a similar form to that of $\sec x$’s derivative. It contains two components: the function itself, $\boldsymbol{\csc x}$, and a second factor, $\boldsymbol{\cot x}$.

\begin{aligned}\dfrac{d}{dx} = – \csc x \cot x\end{aligned}

In the next section, we’ll understand why we have to account for the formula’s negative sign. For now, it’s important to know this derivative rule, and we’ll be using this when we integrate complex expressions in the future.

Through the derivative of $\csc x$, we can now completely differentiate functions that contain $\csc x$ or $\csc[g(x)]$, where $g(x)$ is a differentiable function. Here are some examples of functions that will benefit from the derivative of $\csc x$.

\begin{aligned}f(x) &= \csc (2x)\\ g(x) &= \dfrac{-\sin x + \csc x }{2x}\\h(x) &= \dfrac{\csc(\sqrt{2x + 3})}{x^2 -1}\end{aligned}

Before we apply this rule on functions we’d like to differentiate, let’s go ahead and understand how we came up with the formula for $\dfrac{d}{dx} \csc x$.

How to find the derivative of csc and functions containing it?

The derivative of $\csc x$ and $\sec x $ are so similar that their derivations also follow a similar approach. Yes, we will apply the quotient rule once we’ve rewritten $\csc x$ in terms of $\sin x$.

In the past, we’ve learned that $\csc x = \dfrac{1}{\sin x}$, so we can take the derivative of $\dfrac{1}{\sin x}$ instead to find $\dfrac{d}{dx} \csc x$. We can then apply the quotient rule, $\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{g(x)\dfrac{d}{dx}f(x) – f(x) \dfrac{d}{dx} g(x)}{[g(x)]^2}$, to differentiate the expression.

\begin{aligned}\dfrac{d}{dx} \csc x &=\dfrac{d}{dx} \dfrac{1}{\sin x}\\&= \dfrac{\sin x \dfrac{d}{dx}(1) – 1\dfrac{d}{dx} (\sin x)}{(\sin x)^2}, \phantom{x}\color{Teal}\text{Quotient Rule}\\&= \dfrac{\sin x {\color{Teal}(0)} – 1{\color{Green}(\cos x)}}{(\cos x)^2}, \phantom{x}{\color{Teal}\text{Constant Rule}}\phantom{x}\text{&}{\color{Green}\text{ Derivative of Cosine}}\end{aligned}

We can rewrite this expression using trigonometric identities as shown below.

\begin{aligned}\dfrac{d}{dx} \csc x &= \dfrac{0 – \sin x}{(\sin x)^2}\\&= -\dfrac{1}{\sin x}\cdot\dfrac{\cos x}{\sin x}\\&= -\csc x \cot x\end{aligned}

Thus, we’ve proven the derivative rule for $\csc x$: $\dfrac{d}{dx} \csc x = -\csc x \cot x$.  We can now use this to also differentiate other functions such as $y = \csc (4x – 1)$.

There are instances where we might have to keep the chain rule in mind. The function, $y = \csc (4x – 1)$ is a great example to show you why.

\begin{aligned}h'(x) &= \dfrac{d}{dx} f[g(x)] \\&= f'[g(x)] \cdot g'(x)\\f(x) &= \csc x\\ g(x) &= 4x – 1\\h'(x) &= \csc (4x – 1)\end{aligned}

We can apply the derivative rule for $\csc x$, $\dfrac{d}{dx} \csc  x = -\csc x \cot x$ to find $f’(x)$. We can use other derivative rules to differentiate the inner function, $g(x)$.

\begin{aligned}h'(x)&= {\color{Teal}\csc[g(x)]\tan [g(x)]} \cdot g'(x),\phantom{x} {\color{Teal} \text{Derivative of Cosecant}}\\&= -\csc(4x -1)\cot(4x -1) \cdot g'(x)\\&=-\csc(4x -1)\cot(4x -1)\cdot\color{Teal}\left(\dfrac{d}{dx}4x -\dfrac{d}{dx}1\right),\phantom{x} {\color{Teal} \text{Difference Rule}} \\&=-\csc(4x -1)\cot(4x -1)\cdot\left({\color{Teal}4\dfrac{d}{dx}x}+{\color{Green}0}\right),\phantom{x} {\color{Teal} \text{Constant Multiple Rule }}\text{&}{\color{Green} \text{ Constant Rule}}\\&=-\csc(4x -1)\cot(4x -1)\cdot\left(4({\color{Teal}1})+ 0\right),\phantom{x} {\color{Teal} \text{Power Rule }}\\&=-4\csc(4x -1)\cot(4x -1)\end{aligned}

Hence, $y’ = -4\csc(4x – 1)\cot (4x – 1)$. This shows us that with the help of other derivative rules, we can now differentiate functions that contain cosecant in their expressions. Here are some more examples for you to work on and practice what you’ve just learned.

Example 1

Find the derivative of $h(x) = \csc [(2x – 1)^2]$.


Our given function is a composite function once again. This means that we’ll need to apply the chain rule to account for the inner function, $(2x -1)^2$.

\begin{aligned}h'(x) &= \dfrac{d}{dx} f[g(x)] \\&= f'[g(x)] \cdot g'(x)\\f(x) &= \csc x\\ g(x) &= 2(x-1)^2\\h'(x) &= \csc[2(x – 1)^2]\end{aligned}

We can apply the derivative rule for cosecant to differentiate $f(x)$. We can use the power and chain rules to differentiate $g(x)$ completely.

\begin{aligned}h'(x)&= {\color{Teal}\csc[g(x)]\cot[g(x)]} \cdot g'(x),\phantom{x} {\color{Teal} \text{Derivative of Cosecant}}\\&=- \csc[2(x – 1)^2]\cot[2(x – 1)^2]]\cdot g'(x)\\&=\csc[2(x – 1)^2]\cot[2(x – 1)^2]\cdot\left[{\color{Teal}2(2x- 1)^{2 – 1}}\cdot {\color{Green}\dfrac{d}{dx}(2x -1)} \right],\phantom{x} {\color{Teal} \text{Power Rule }}\text{&}{\color{Green} \text{ Chain Rule}}\\&= \csc[2(x – 1)^2]\cot[2(x – 1)^2]\cdot\left[2(2x- 1)\cdot \left({\color{Teal}2\dfrac{d}{dx}x – \dfrac{d}{dx}1 }\right )\right],\phantom{x} {\color{Teal} \text{Difference Rule }}\end{aligned}

Apply a few more derivative rules to simplify $h’(x)$ further.

\begin{aligned}h'(x)&=- \csc[2(x – 1)^2]\tan[2(x – 1)^2]\cdot\left[2(2x- 1)\cdot (2({\color{Teal}1 }) – {\color{Green}0})\right],\phantom{x} {\color{Teal} \text{Power Rule }}\text{&}{\color{Green} \text{ Constant Rule}}\\&=-\csc[2(x – 1)^2]\tan[2(x – 1)^2]\cdot[4(2x – 1)]\\&=-(8x- 4)\csc[2(x – 1)^2]\tan[2(x – 1)^2] \end{aligned}

This shows that with fundamental derivative rules and our knowledge of $\csc x$’s derivative, we have $y’ =-(8x- 4)\csc[2(x – 1)^2]\tan[2(x – 1)^2]$.

Example 2

Find the derivative of $h(x) = \dfrac{e^x }{\csc (3x + 5)}$.


Since $h(x)$ is a rational expression, we’ll need the quotient rule to differentiate $h(x)$.

\begin{aligned}h'(x)&= \dfrac{[\csc (3x + 5)]\dfrac{d}{dx}e^x – e^x \dfrac{d}{dx} \csc(3x +5)}{[\csc(3x +5)]^2}\end{aligned}

Recall that $\dfrac{d}{dx} e^x$ is simply $e^x$, so we can use this to simplify the first group in the numerator. We can use the derivative of $\csc x$ and the chain rule to simplify the second group of terms in the numerator.

\begin{aligned}h'(x)&= \dfrac{{\color{Teal}e^x}[\csc (3x + 5)] – e^x \dfrac{d}{dx} \csc(3x +5)}{[\csc(3x +5)]^2},\phantom{x}\color{Teal}\text{Derivative of }e^x\\&=\dfrac{e^x[\csc (3x + 5)] – e^x[{\color{Teal}-\csc(3x +5)\cot(3x +5) }]\cdot {\color{Green}\dfrac{d}{dx}(3x + 5)}}{[\csc(3x +5)]^2},\phantom{x}\color{Teal}\text{Derivative of Cosecant }\text{&}\color{Green}\text{ Chain Rule} \end{aligned}

Let’s continue to simplify this expression using the derivative rules we’ve learned in the past.

\begin{aligned}h'(x)&=\dfrac{e^x[\csc (3x + 5)] +e^x[\csc(3x +5)\cot(3x +5) ]\cdot {\color{Teal}\left(\dfrac{d}{dx}3x + \dfrac{d}{dx}5\right)}}{[\csc(3x +5)]^2},\phantom{x}\color{Teal}\text{ Sum Rule}\\&=\dfrac{e^x[\csc (3x + 5)] +e^x[\csc(3x +5)\cot(3x +5) ]\cdot \left({\color{Teal}3\dfrac{d}{dx}x }+ {\color{Green}0}\right)}{[\csc(3x +5)]^2},\phantom{x}\color{Teal}\text{ Constant Multiple Rule }\text{&}\color{Green}\text{ Constant Rule}\\&=\dfrac{e^x[\csc (3x + 5)] +e^x[\csc(3x +5)\cot(3x +5) ]\cdot \left(3{\color{Teal}(1)}\right)}{[\csc(3x +5)]^2},\phantom{x}\color{Teal}\text{ Power Rule }\\&= \dfrac{e^x[\csc (3x + 5)] +3e^x[\csc(3x +5)\cot(3x +5) ]}{[\csc(3x +5)]^2}\\&= \dfrac{e^x[1+ 3\cot(3x +5)]}{\csc(3x + 5)}\end{aligned}

Hence, through the derivative of $\csc x$, we have $h’(x)=  \dfrac{e^x[1+ 3\cot(3x +5)]}{\csc(3x + 5)}$.

Practice Questions

1. Find the derivative of $f(x) = -4\csc (2x +1)$.
2. Find the derivative of $g(x) = 2\csc (\sqrt{3x – 1})$.
3. Find the derivative of $h(x) = e^x[\csc(x^2 + 1)]$.
4. Find the derivative of $y = \dfrac{\csc (5x)}{\csc(2x + 1)}$.
5. Find the derivative of $y = \dfrac{e^x \csc x}{\csc(3x^2 + 1)}$.

Answer Key

1. $f’(x) =8\csc(2x + 1)\cot(2x +1)$
2. $g’(x) = -\dfrac{3\csc(\sqrt{3} – 1)(\cot \sqrt{3} – 1)}{\sqrt{3x – 1}} $
3. $h’(x) = -e^x \csc(x^2 + 1)[2x\cot(x^2 + 1) – 1]$
4. $y’ = 5\cot \left(5x\right)\csc \left(5x\right)\sin \left(2x+1\right)+2\cos \left(2x+1\right)\csc \left(5x\right)$
5. $y’= -\cot \left(x\right)\csc \left(x\right)\sin \left(3x^2+1\right)+6x\cos \left(3x^2+1\right)\csc \left(x\right)$