**The derivative of the inverse cosecant function is equal to -1/(|x|√(x ^{2}-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.

##### CALCULUS

**Relevant for**…

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

## Avoid confusion in using the denotations arccsc(x), csc^{-1}(x),
1
/
csc(x) , and csc^{n}(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 $latex f(x) = \csc^{-1}(6x)$

##### Solution

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 $latex F(x) = \csc^{-1}(x^3-8)$?

##### Solution

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

##### Solution

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

## See also

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

- Derivative of arcsin (Inverse Sine) With Proof and Graphs
- Derivative of arccos (Inverse Cosine) With Proof and Graphs
- Derivative of arctan (Inverse Tangent) With Proof and Graphs
- Derivative of arcsec (Inverse Secant) With Proof and Graphs
- Derivative of arccot (Inverse Cotangent) With Proof and Graphs

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