# Derivative of Cotangent Squared, cot^2(x) with Proof and Graphs

The derivative of cotangent squared is equal to minus two cotangent times cosecant squared, -2cot(x)csc2(x). This derivative can be found using the chain rule and the derivatives of the fundamental trigonometric functions.

Here, we will learn how to prove this derivative, we will look at a graphical comparison of the cotangent squared and its derivative, and we will solve some exercises.

##### CALCULUS

Relevant for

Learning how to find the derivative of cotangent squared.

See proof

##### CALCULUS

Relevant for

Learning how to find the derivative of cotangent squared.

See proof

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

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

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

Because this is a composite function, the chain rule formula is used as a more direct tool to prove the derivative of the cotangent squared function.

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

For a better representation, we can rewrite it as

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

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

It becomes evident that the given power function is the outer function to be considered, while the cotangent function, being raised by the given power function, is the inner function. We can configure the outer function as follows:

$latex f(u) = u^2$

where

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

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

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

$latex u = g(x)$

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

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 cotangent in terms of x, we have

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

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

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

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

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

Substituting u into f'(u), we have

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

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

which brings us to the derivative formula of cosecant squared x

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

## Why are the derivative of the cotangent squared and the cosecant squared the same?

You may wonder why the derivative of both functions

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

and

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

are the same.

The Pythagorean trigonometric identity for cotangents and cosecants tells us that

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

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

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

Now, we see that on the right-hand side, we have a derivative of a constant, which is equal to zero. Then,

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

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

This is why both the cotangent squared and the cosecant squared have the same derivative.

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

The graph of the function

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

is

By deriving $latex f(x) = \cot^{2}{(x)}$, we get

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

and its graph is

Comparing their graphs, we have:

Using the graphs, we can see that the original function $latex f(x) = \cot^{2}{(x)}$ 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

$latex [0,\infty)$

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

The following examples show how to find the derivative of the composite cotangent squared.

### EXAMPLE 1

Determine the derivative of the function $latex f(x) = \cot^2(7x)$.

To derive this function, we have to use the chain rule, since it is a composite cotangent squared function.

Therefore, if we consider $latex u=7x$ as the inner function, we can write $latex f(u)=\cot^2(u)$. Using the chain rule, we have:

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

$$\frac{dy}{dx}=-2\cot(u)\csc^2(u) \times 7$$

Now, we substitute $latex u=7x$ back into the function and we have:

$$\frac{dy}{dx}=-14\cot(7x)\csc^2(7x)$$

### EXAMPLE 2

What is the derivative of the function $latex F(x) = \cot^2(3x^3-5x)$?

Here, the inner function is $latex u=3x^3-5x$. Therefore, the outer function is $latex f (u) = \cot^2(u)$.

Now, we start by finding the derivative of the external function:

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

Then, we find the derivative of the inner function and we have:

$$\frac{d}{dx}(g(x)) = \frac{d}{dx}(3x^3-5x)$$

$$\frac{d}{dx}(g(x)) = 9x^2-5$$

By the chain rule, we multiply the derivative of $latex f(u)$ by the derivative of $latex g(x)$ and we have:

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

$$\frac{dy}{dx} = -2\cot(u)\csc^2(u) \cdot (9x^2-5)$$

Finally, we apply the substitution $latex u=3x^3-5x$ and simplify:

$$\frac{dy}{dx} = -2\cot(3x^3-5x)\csc^2(3x^3-5x) \cdot (9x^2-5)$$

$$\frac{dy}{dx} = -(18x^2-10)\cot(3x^3-5x)\csc^2(3x^3-5x)$$

## Practice of derivatives of cotangent squared functions

Derivatives of cotangent squared quiz
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