# Derivative of arccsc (Inverse Cosecant) With Proof and Graphs

The derivative of the inverse cosecant function is equal to -1/(|x|√(x2-1)). This derivative can be derived using the Pythagorean theorem and Algebra.

In this article, we will learn how to derive the inverse cosecant function. We will see brief fundamentals, a proof, a graphical comparison of the function and its derivative, and some examples.

##### CALCULUS

Relevant for

Learning about the proof and graphs of the derivative of arccsc of x.

See proof

##### CALCULUS

Relevant for

Learning about the proof and graphs of the derivative of arccsc of x.

See proof

## Avoid confusion in using the denotations arccsc(x), csc-1(x), 1 / csc(x) , and cscn(x)

The use of the different denotations $latex \text{arccsc}(x)$, $latex \csc^{-1}{(x)}$, $latex \frac{1}{\csc{(x)}}$ y $latex \csc^{n}{(x)}$ can cause some confusion. It is important not to interchange the meaning of these symbols, as it can lead to derivation errors.

Summarizing the definition of these symbols, we have

$latex \text{arccsc}(x) = \csc^{-1}{(x)}$

Both the symbols $latex \text{arccsc}$ and $latex \csc^{-1}$ are used to represent the inverse cosecant. $latex \text{arccos}$ is commonly used as the verbal symbol for the inverse cosecant function, while $latex \csc^{-1}$ is used as the mathematical symbol for the inverse cosecant function for a more formal setting.

In the case of the denotation $latex \csc^{-1}{(x)}$, we must consider that $latex -1$ is not an algebraic exponent of a cosecant. The $latex -1$ used for inverse cosecant represents that the cosecant is inverse and not raised to $latex -1$.

Therefore,

$latex \csc^{-1}{(x)} \neq \frac{1}{\csc{(x)}}$

And givens such as $latex \csc^{2}{(x)}$ or $latex \csc^{n}{(x)}$, where n is any algebraic exponent of a non-inverse cosecant, MUST NOT use the inverse cosecant formula since in these givens, both the 2 and any exponent n are treated as algebraic exponents of a non-inverse cosecant.

## Proof of the Derivative of the Inverse Cosecant Function

In this proof, we will mainly use the concepts of a right triangle, the Pythagorean theorem, the trigonometric functions of cosecant and cotangent, and some basic algebra. Just like in the previous figure as a reference sample for a given right triangle, suppose we have that same triangle $latex \Delta ABC$, but this time, let’s change the variables for an easier illustration.

where for every one-unit of a side opposite to angle y, there is a side $latex \sqrt{x^2-1}$ adjacent to angle y and a hypothenuse x.

Using these components of a right-triangle, we can find the angle y by using Cho-Sha-Cao, particularly the cosecant function by using the hypothenuse x and its opposite side.

$latex \csc{(\theta)} = \frac{hyp}{opp}$

$latex \csc{(y)} = \frac{x}{1}$

$latex \csc{(y)} = x$

Now, we can implicitly derive this equation by using the derivative of trigonometric function of cosecant for the left-hand side and power rule for the right-hand side. Doing so, we have

$latex \frac{d}{dx} (\csc{(y)}) = \frac{d}{dx} (x)$

$latex \frac{d}{dx} (\csc{(y)}) = 1$

$latex \frac{dy}{dx} (-\csc{(y)}\cot{(y)}) = 1$

$latex \frac{dy}{dx} = \frac{1}{-\csc{(y)}\cot{(y)}}$

$latex \frac{dy}{dx} = -\frac{1}{\csc{(y)}\cot{(y)}}$

Getting the tangent of angle y from our given right-triangle, we have

$latex \cot{(y)} = \frac{adj}{opp}$

$latex \cot{(y)} = \frac{\sqrt{x^2-1}}{1}$

$latex \cot{(y)} = \sqrt{x^2-1}$

We can then substitute $latex \csc{(y)}$ and $latex \cot{(y)}$ to the implicit differentiation of $latex \csc{(y)} = x$

$latex \frac{dy}{dx} = -\frac{1}{\csc{(y)}\cot{(y)}}$

$latex \frac{dy}{dx} = -\frac{1}{(x) \cdot \left(\sqrt{x^2-1}\right)}$

$latex \frac{dy}{dx} = -\frac{1}{x\sqrt{x^2-1}}$

Now, since

$latex \csc{(y)} = x$

and

$latex hypothenuse = x$

We know that a negative hypothenuse cannot exist. Therefore, $latex \csc{(y)}$ in this case cannot be negative. That’s why the x multiplicand in the denominator of the derivative of inverse cosecant must be considered an absolute value.

$latex \frac{dy}{dx} = -\frac{1}{|x|\sqrt{x^2-1}}$

Therefore, algebraically solving for the angle y and getting its derivative, we have

$latex \csc{(y)} = x$

$latex y = \frac{x}{\csc}$

$latex y = \csc^{-1}{(x)}$

$latex \frac{dy}{dx} = \frac{d}{dx} \left( \csc^{-1}{(x)} \right)$

$latex \frac{dy}{dx} = -\frac{1}{|x|\sqrt{x^2-1}}$

which is now the derivative formula for the inverse cosecant of x.

Now, for the derivative of an inverse cosecant of any function other than x, we may apply the derivative formula of inverse cosecant together with the chain rule formula. By doing so, we have

$latex \frac{dy}{dx} = \frac{d}{du} \csc^{-1}{(u)} \cdot \frac{d}{dx} (u)$

$latex \frac{dy}{dx} = -\frac{1}{|u|\sqrt{u^2-1}} \cdot \frac{d}{dx} (u)$

where $latex u$ is any function other than x.

## Graph of Inverse Cosecant x VS. The Derivative of Inverse Cosecant x

Given the function

$latex f(x) = \csc^{-1}{(x)}$

its graph is

And as we know by now, by deriving $latex f(x) = \csc^{-1}{(x)}$, we get

$latex f'(x) = -\frac{1}{|x|\sqrt{x^2-1}}$

which has the graph as

Illustrating both graphs in one, we have

Analyzing these graphs, it can be seen that the original function $latex f(x) = \csc^{-1}{(x)}$ has a domain of

$latex (-\infty,-1] \cup [1,\infty)$ or all real numbers except $latex -1 < x < 1$

and exists within the range of

$latex \left[-\frac{\pi}{2},0\right) \cup \left(0,\frac{\pi}{2}\right]$ or $latex -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$ except zero

whereas the derivative $latex f'(x) = -\frac{1}{|x|\sqrt{x^2-1}}$ has a domain of

$latex (-\infty,-1) \cup (1,\infty)$ or all real numbers except $latex -1 \leq x \leq 1$

and exists within the range of

$latex (-\infty,0)$ or $latex y < 0$

## Examples

The following examples show how to derive composite inverse cosecant functions.

### EXAMPLE 1

Find the derivative of $latex f(x) = \csc^{-1}(6x)$

To derive this function we use the chain rule since we have a composite cosecant function.

We start by considering $latex u=6x$ as the inner function. This means that we have $latex f(u)=\csc^{-1}(u)$ and using the chain rule, we have:

$$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$$

$$\frac{dy}{dx}=-\frac{1}{|u|\sqrt{u^2-1}} \times 6$$

Now, we just have to substitute $latex u=6x$ back into the function and we have:

$$\frac{dy}{dx}=-\frac{6}{|6x|\sqrt{(6x)^2-1}}$$

$$\frac{dy}{dx}=-\frac{6}{|6x|\sqrt{36x^2-1}}$$

### EXAMPLE 2

What is the derivative of the function $latex F(x) = \csc^{-1}(x^3-8)$?

We are going to use the chain rule. Therefore, we write $latex f (u) = \csc^{-1}(u)$, where $latex u = x^3-8$.

Now, we calculate the derivative of the outer function $latex f(u)$:

$$\frac{d}{du} ( \csc^{-1}(u) ) = -\frac{1}{|u|\sqrt{u^2-1}}$$

Then, we determine the derivative of the inner function $latex g(x)=u=x^3-8$:

$$\frac{d}{dx}(g(x)) = \frac{d}{dx}(x^3-8)$$

$$\frac{d}{dx}(g(x)) = 3x^2$$

Then, we multiply the derivative of the outer function by the derivative of the inner function:

$$\frac{dy}{dx} = \frac{d}{du} (f(u)) \cdot \frac{d}{dx} (g(x))$$

$$\frac{dy}{dx} = -\frac{1}{|u|\sqrt{u^2-1}} \cdot 3x^2$$

As a last step, we substitute $latex u=x^3-8$ back in and simplify:

$$\frac{dy}{dx} = -\frac{1}{|x^3-8|\sqrt{(x^3-8)^2-1}} \cdot 3x^2$$

$$\frac{dy}{dx} = -\frac{3x^2}{|x^3-8|\sqrt{(x^3-8)^2-1}}$$

$$F'(x) = -\frac{3x^2}{|x^3-8|\sqrt{x^6-16x^3+63}}$$

### EXAMPLE 3

What is the derivative of $latex f(x) = \csc^{-1}(\sqrt{x})$?

The internal function of the inverse cosecant is $latex u=\sqrt{x}$. Since we can write it as $latex u=x^{\frac{1}{2}}$, its derivative is:

$$\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}$$

Applying the chain rule with $latex f(u)=\csc^{-1}(u)$, we have:

$$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$$

$$\frac{dy}{dx}=-\frac{1}{|u|\sqrt{u^2-1}} \times \frac{1}{2}x^{-\frac{1}{2}}$$

Substituting $latex u=\sqrt{x}$ back and simplifying, we have:

$$\frac{dy}{dx}=-\frac{1}{|\sqrt{x}|\sqrt{(\sqrt{x})^2-1}} \times \frac{1}{2}x^{-\frac{1}{2}}$$

$$\frac{dy}{dx}=-\frac{1}{|\sqrt{x}|\sqrt{x-1}} \times \frac{1}{2}x^{-\frac{1}{2}}$$

$$\frac{dy}{dx}=-\frac{1}{2|\sqrt{x}|\sqrt{x-1}\sqrt{x}}$$

$$\frac{dy}{dx}=-\frac{1}{2|\sqrt{x}|\sqrt{x(x-1)}}$$

## Practice of derivatives of composite inverse cosecant functions

Derivatives of inverse cosecant quiz  You have completed the quiz!

Interested in learning more about the derivatives of trigonometric functions? Take a look at these pages: ### Jefferson Huera Guzman

Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.  