# Derivative of Cosecant Squared, csc^2(x) with Proof and Graphs

The derivative of cosecant squared is equal to minus two contingent times cosecant squared, -2cot(x)csc2(x). We can find this derivative by using the chain rule and the derivatives of the fundamental trigonometric functions.

Here, we will look at a demonstration of this derivative, the graphical comparison of the cosecant squared and its derivative, and some examples.

##### CALCULUS

Relevant for

Learning how to find the derivative of cosecant squared.

See proof

##### CALCULUS

Relevant for

Learning how to find the derivative of cosecant squared.

See proof

## Proof of The Derivative of Cosecant Squared Function with the use of Chain Rule

As a prerequisite, please review the chain rule formula and its proof by looking at the article: Chain Rule of derivatives. Similarly, you can review the proof of the derivative of cosecant function by visiting this article: Derivative of Cosecant, csc(x).

$latex \csc^{2}{(x)} \neq \csc{(x^2)}$

This is a composite cosecant function, so it can be differentiated using the chain rule. Therefore, we start with the function:

$latex F(x) = \csc^{2}{(x)}$

We can identify the two functions that make up F(x). One function is raised to a power of two and the other is a trigonometric function of cosecant.

If we rewrite the function as follows, we can see this more clearly:

$latex F(x) = \csc^{2}{(x)}$

$latex F(x) = (\csc{(x)})^2$

Now it is clear that the power function is the external function, while the cosecant function is the internal function. We can configure the outer function as follows:

$latex f(u) = u^2$

where

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

The trigonometric cosecant function, as the inner function of f(u), will be denoted as g(x).

$latex f(u) = f(g(x))$

$latex u = g(x)$

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

Deriving the outer function f(u) using the power rule in terms of u, we have

$latex f(u) = u^2$

$latex f'(u) = 2u$

Deriving the inner function g(x) using the derivative formula of trigonometric function cosecant in terms of x, we have

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

$latex g'(x) = -\csc{(x)}\cot{(x)}$

Algebraically multiplying the derivative of outer function $latex f'(u)$ by the derivative of inner function $latex g'(x)$, we have

$latex \frac{dy}{dx} = f'(u) \cdot g'(x)$

$latex \frac{dy}{dx} = (2u) \cdot (-\csc{(x)}\cot{(x)})$

Substituting u into f'(u), we have

$latex \frac{dy}{dx} = (2(\csc{(x)})) \cdot (-\csc{(x)}\cot{(x)})$

$latex \frac{dy}{dx} = -(2(\csc{(x)})) \cdot (\csc{(x)}\cot{(x)})$

$latex \frac{dy}{dx} = -2\csc^{2}{(x)} \cdot \cot{(x)}$

which brings us to the derivative formula of cosecant squared x

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

## Relationship between the derivative of cosecant squared and cotangent squared

You may wonder why

$latex \csc^{2}{(x)}$

and

$latex \cot^{2}{(x)}$

have similar derivatives.

According to the Pythagorean formula for cosecants and cotangents,

$latex \csc^{2}{(x)} = 1 + \cot^{2}{(x)}$

If we try to derive both sides of the equation, we have

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

Evaluating the derivative of the first term in the right-hand-side of the equation, where the derivative is zero, we have

$latex \frac{d}{dx} (\csc^{2}{(x)}) = 0 + \frac{d}{dx}(\cot^{2}{(x)})$

$latex \frac{d}{dx} (\csc^{2}{(x)}) = \frac{d}{dx}(\cot^{2}{(x)})$

This is why both the cosecant squared and the cotangent squared have the same derivative, because of the Pythagorean formula for cosecants and cotangents.

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

Given the function

$latex f(x) = \csc^{2}{(x)}$

it is graphed as

As we already know, deriving $latex f(x) = \csc^{2}{(x)}$ is

$latex f'(x) = -2\cot{(x)}\csc^{2}{(x)}$

which is graphed as

Illustrating both graphs in one, we have

By examining the differences between these functions basing on the graphs above, you can see that the original function $latex f(x) = \csc^{2}{(x)}$ has a domain of

$$(-2\pi,-\pi) \cup (-\pi,0) \cup (0,\pi) \cup (\pi,2\pi)$$

within the finite intervals of

$latex (-2\pi,2\pi)$

and lies within the range of

$latex [1,\infty)$

whereas the derivative $latex f'(x) = -2\cot{(x)}\csc^{2}{(x)}$ has a domain of

$$(-2\pi,-\pi) \cup (-\pi,0) \cup (0,\pi) \cup (\pi,2\pi)$$

within the finite intervals of

$latex (-2\pi,2\pi)$

and lies within the range of

$latex (-\infty,\infty)$

## Examples

In the following examples, we can see how to derive a compound cosecant squared function.

### EXAMPLE 1

Determine the derivative of the function $latex f(x) = \csc^2(15x)$.

This is a composite cosecant squared function, so the use of the chain rule is required to find its derivative.

Therefore, let’s use $latex u=15x$ as the inner function, so we have $latex f(u)=\csc^2(u)$. Using the chain rule, this gives us:

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

$$\frac{dy}{dx}=-2\cot(u)\csc^2(u) \times 15$$

We just have to substitute $latex u=15x$ back into the function and we have:

$$\frac{dy}{dx}=-30\cot(15x)\csc^2(15x)$$

### EXAMPLE 2

Find the derivative of the function $latex F(x) = \csc^2(6x^3-8x)$

In this case, the inner function is $latex 6x^3-8x$, which allows us to write $latex f (u) = \csc^2(u)$.

Then, we can find the derivative of the external function

$$\frac{d}{du} ( \csc^2(u) ) = -2\cot(u)\csc^2(u)$$

Calculating the derivative of $latex g(x)=u=6x^3-8x$, we have:

$$\frac{d}{dx}(g(x)) = \frac{d}{dx}(6x^3-8x)$$

$$\frac{d}{dx}(g(x)) = 18x^2-8$$

Multiplying the derivative of the external function $latex f(u)$ by the derivative of the internal function $latex g(x)$, we have:

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

$$\frac{dy}{dx} = -2\cot(u)\csc^2(u) \cdot (18x^2-8)$$

Finally, we apply the substitution $latex u=6x^3-8x$ and simplify:

$$\frac{dy}{dx} = -2\cot(6x^3-8x)\csc^2(6x^3-8x) \cdot (18x^2-8)$$

$$\frac{dy}{dx} = -(36x^2-16)\cot(6x^3-8x)\csc^2(6x^3-8x)$$

## Practice of derivatives of cosecant squared functions

Derivatives of cosecant squared quiz
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