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

The cotangent squared function is the cotangent raised to the power of two. 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.

We’ll go through the basics, definition, formula, graph comparison of underived and derived cotangent squared, proof, derivation techniques, and some more relevant instances.

##### 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).

Please be reminded that

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

To prevent misunderstanding, the first is a “full trigonometric function” raised to the power of two, whereas the second is a trigonometric function of “the square of a variable.”

Because it is a composite function, the chain rule formula is used as a more straightforward tool to prove the derivative of the cotangent squared function, providing you already know how to prove the chain rule formula and the derivative of a cotangent function.

Assuming we are asked to find the derivative of

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

We can identify the two functions that make up F(x). There is a power function and a trigonometric function in this scenario. To be more exact, these are a function raised to a power of two and a trigonometric function of cotangent, based on our given F(x).

For a better representation, we can rewrite it as

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

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

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

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

## Relationship between the derivative of cotangent squared and cosecant squared, why are they the same?

You may wonder why the derivative of both functions

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

and

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

are the same.

According to the Pythagorean formula for cotangents and cosecants,

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

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, which is the derivative of a constant 1, 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 cotangent squared and the cosecant squared have the same derivative.

## How is a Cotangent Squared Function derived? Quicker Methods

Cotangent squared, as previously stated, is a composite function of power and the trigonometric function cotangent. Instead of using the chain rule formula repetitively like what we did in the proof, we may simply use the established derivative formula for a cotangent squared function.

### METHOD 1: Derivative of the square of a cotangent of any angle x in terms of the same angle x.

Step 1: Determine whether the cotangent squared of an angle is a function of the same angle. For instance, if the right-hand side of the equation is $latex \cot^{2}{(x)}$, determine if it is a function of the same angle x or f(x).

Note: If $latex \cot^{2}{(x)}$ is a function of a different angle or variable, such as f(t) or f(y), implicit differentiation will be used, which is beyond the scope of this article.

Step 2: Then directly apply the proven derivative formula of the cotangent squared function

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

If nothing else can be simplified, then that is the final solution.

### METHOD 2: Derivative of the square of a cotangent of any function v in terms of x.

Step 1: Express the function as $latex G(x) = \cot^{2}{(v)}$, where $latex v$ represents any function other than x.

Step 2: Treat $latex \cot^{2}{(v)}$ as the outside function $latex g(v)$ and $latex v$ as the inner function $latex h(x)$ of the composite function $latex G(x)$. Doing this, we have

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

and

$latex h(x) = v$

Step 3: Derive the outer function $latex g(v)$, and use the derivative of the cotangent squared function, in terms of $latex v$.

$latex \frac{d}{du} \left( \cot^{2}{(v)} \right) = -2\cot{(v)}\csc^{2}{(v)}$

Step 4: Derive the inner function $latex h(x) = v$. Use whatever appropriate derivative rule applies to $latex v$.

Step 5: Algebraically multiply the derivative of outer function $latex g(v)$ by the derivative of inner function $latex h(x)$ to completely apply chain rule

$latex \frac{dy}{dx} = \frac{d}{du} (g(v)) \cdot \frac{d}{dx} (h(x))$

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

Step 6: Substitute $latex v$ into $latex g'(v)$

Step 7: Simplify and apply any function law whenever applicable then finalize the answer.

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

With the function

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

it is graphed as

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

Here are some examples of deriving a cotangent squared function using either the first or second method.

### EXAMPLE 1

Derive: $latex f(\beta) = \cot^{2}{(\beta)}$

Solution: After examining the given cotangent squared function, it shows that it is only a square of a cotangent of a single angle $latex \beta$. Hence, we can apply the first method to this problem.

Step 1: Assess if the square of cotangent $latex \beta$ is a function of $latex \beta$. It is in this problem. Hence, proceed to step 2.

Step 2: Directly apply the derivative formula of the cotangent squared function and derive in terms of $latex \beta$. Since no further simplification is needed, the final answer is:

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

### EXAMPLE 2

Derive: $latex G(x) = \cot^{2}{(9-4x^2)}$

Solution: After examining the given cotangent squared function, it shows that it is a square of a cotangent of a polynomial function. Hence, we can apply the second method to this problem.

Step 1: Express the cotangent squared function as $latex G(x) = \cot^{2}{(v)}$, using $latex v$ to represent any function other than x, which is the angle of the cotangent squared. In this problem,

$latex v = 9-4x^2$

Let’s substitute this later as we evaluate the derivative of the problem.

Step 2: Consider $latex \cot{(v)}$ as the outside function $latex g(v)$. Then $latex v$ will be the inner function, denoted as $latex h(x)$ too. For this problem, we have

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

and

$latex h(x) = v = 9-4x^2$

Step 3: Derive the outer function $latex g(v)$ using the derivative of the cotangent squared function, in terms of $latex v$.

$latex \frac{d}{du} \left( \cot^{2}{(v)} \right) = -2\cot{(v)}\csc^{2}{(v)}$

Step 4: Derive the inner function $latex h(x)$ or $latex v$. Since our $latex v$ in this problem is a polynomial function, let’s apply the power rule and sum/difference of derivatives to derive $latex v$.

$latex \frac{d}{dx}(h(x)) = \frac{d}{dx} \left(9-4x^2 \right)$

$latex \frac{d}{dx}(h(x)) = -8x$

Step 5: Evaluate the basic chain rule formula by algebraically multiplying the derivative of outer function $latex g(v)$ by the derivative of inner function $latex h(x)$

$latex \frac{dy}{dx} = \frac{d}{du} (g(v)) \cdot \frac{d}{dx} (h(x))$

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

Step 6: Substitute $latex v$ into $latex g'(v)$

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

$$\frac{dy}{dx} = (-2\cot{(9-4x^2)}\csc^{2}{(9-4x^2)}) \cdot (-8x)$$

Step 7: Simplify and apply any function law whenever applicable to finalize the answer.

$$\frac{dy}{dx} = (-2\cot{(9-4x^2)}\csc^{2}{(9-4x^2)}) \cdot (-8x)$$

$$\frac{dy}{dx} = 2\cot{(9-4x^2)}\csc^{2}{(9-4x^2)} \cdot 8x$$

$$\frac{dy}{dx} = 16x \cot{(9-4x^2)}\csc^{2}{(9-4x^2)}$$

And the final answer is:

$$G'(x) = 16x \cot{(9-4x^2)}\csc^{2}{(9-4x^2)}$$  