Derivative of arccot (Inverse Cotangent) With Proof and Graphs

The Derivative of ArcCotagent or Inverse Cotangent is used in deriving a function that involves the inverse form of the trigonometric function ‘cotangent‘. The derivative of the inverse cotangent function is equal to -1/(1+x2). This derivative can be proved using the Pythagorean theorem and Algebra.

In this article, we will discuss how to derive the arccotangent or the inverse cotangent function. We will cover brief fundamentals, its definition, formula, a graph comparison of the underived and derived function, a proof, methods to derive, and a few examples.

CALCULUS
Derivative of arccot, inverse cotangent

Relevant for

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

See proof

CALCULUS
Derivative of arccot, inverse cotangent

Relevant for

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

See proof

Avoid confusion in using the denotations arccot(x), cot-1(x),
1 / cot(x)
, and cotn(x)

It is important that we do not compromise the potential confusions we might have in using different denotations between $latex \text{arccot}(x)$, $latex \cot^{-1}{(x)}$, $latex \frac{1}{\cot{(x)}}$, and $latex \cot^{n}{(x)}$, since interchanging the meaning of these symbols may lead to derivation mistakes. Summarizing the definition of these symbols, we have

$latex \text{arccot}(x) = \cot^{-1}{(x)}$

Both symbols $latex \text{arccot}$ and $latex \cot^{-1}$ can be used interchangeably when computing for the inverse cotangent of either a variable or another function. \text{arccot} is commonly used as the verbal symbol of inverse cotangent function which is popularly used as introductory denotations for beginners whilst $latex \cot^{-1}$ is used as a mathematical symbol of inverse cotangent function for a more professional setting.

However, when it comes to the denotation $latex \cot^{-1}{(x)}$, sometimes it can confuse learners that $latex -1$ is an algebraic exponent of a non-inverse cotangent, which is not true. The $latex -1$ used for inverse cotangent represents the cotangent being inverse and not raised to $latex -1$. This has been proven and shown in the previous sub-article written above.

Therefore,

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

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


Proof of the Derivative of the Inverse Cotangent Function

In this proof, we will mainly use the concepts of a right triangle, the Pythagorean theorem, the trigonometric functions of cotangent and cosecant, 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.

Right-Triangle-coty-x

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

Using these components of a right-triangle, we can find the angle y by using Cho-Sha-Cao, particularly the cotangent function by using its adjacent and opposite sides.

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

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

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

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

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

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

$latex \frac{dy}{dx} (-\csc^{2}{(y)}) = 1$

$latex \frac{dy}{dx} = \frac{1}{-\csc^{2}{(y)}}$

$latex \frac{dy}{dx} = -\frac{1}{\csc^{2}{(y)}}$

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

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

$latex \csc{(y)} = \frac{\sqrt{1+x^2}}{1}$

Since we need to substitute $latex \csc{(y)}$ into $latex \csc^{2}{(y)}$, we need to square both sides

$latex \csc^{2}{(y)} = \left(\frac{\sqrt{1+x^2}}{1}\right)^2$

$latex \csc^{2}{(y)} = \left(\sqrt{1+x^2}\right)^2$

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

We can then substitute $latex \csc^{2}{(y)}$ to the implicit differentiation of $latex \cot{(y)} = x$

$latex \frac{dy}{dx} = -\frac{1}{\csc^{2}{(y)}}$

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

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

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

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

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

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

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

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

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

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

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

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


How to derive an Inverse Cotangent Function?

The derivative process of an inverse cotangent function is very straightforward assuming you have already learned the concepts behind the usage of the inverse cotangent function and how we arrived to its derivative formula.

METHOD 1: Derivative of the Inverse Cotangent of any single variable x

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

Step 1: Analyze if the inverse cotangent of a variable is a function of that same variable. For example, if the right-hand side of the equation is $latex \cot^{-1}{(x)}$, then check if it is a function of the same variable x or f(x).

Note: If $latex \cot^{-1}{(x)}$ is a function of a different variable such as f(t) or f(y), it will use implicit differentiation which is out of the scope of this article.

Step 2: Illustrate derivation by using the derivation symbols or denotations such as

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

or

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

Step 3: Then directly apply the derivative formula of inverse cotangent function

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

If nothing is to be simplified anymore, then that would be the final answer.

METHOD 2: Derivative of the Inverse Cotangent of any function u in terms of x.

$latex \frac{d}{dx} \left( \cot^{-1}{(u)} \right) = -\frac{1}{1+u^2} \cdot \frac{d}{dx} (u)$

Step 1: Express the inverse function as $latex F(x) = \cot^{-1}{(u)}$ or $latex F(x) = \text{arccot}(u)$, where $latex u$ represents any function other than x.

Step 2: Illustrate derivation by using the derivation symbols or denotations such as

$latex \frac{d}{dx} F(x) = \frac{d}{dx} \left(\cot^{-1}{(u)} \right)$

or

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

Step 3: Consider $latex \cot^{-1}{(u)}$ as the outside function $latex f(u)$ and $latex u$ as the inner function $latex g(x)$ of the composite function $latex F(x)$. Hence we have

$latex f(u) = \cot^{-1}{(u)}$

and

$latex g(x) = u$

Step 4: Get the derivative of the outer function $latex f(u)$, which must use the derivative of inverse cotangent $latex \cot^{-1}{(u)}$, in terms of $latex u$.

$latex \frac{d}{du} \left( \cot^{-1}{(u)} \right) = -\frac{1}{1+u^2}$

Step 5: Get the derivative of the inner function $latex g(x) = u$. Use the appropriate derivative rule that applies to $latex u$.

Step 6: Apply the basic chain rule formula by algebraically multiplying the derivative of outer function $latex f(u)$ by the derivative of inner function $latex g(x)$

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

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

Step 7: Substitute $latex u$ into $latex f'(u)$

Step 8: Simplify and apply any function law whenever applicable to finalize the answer.


Graph of Inverse Cotangent x VS. The Derivative of Inverse Cotangent x

Given the function

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

the graph is illustrated as

graph-of-arccotx

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

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

which is illustrated graphically as

graph of derivative of arccotx

Illustrating both graphs in one, we have

graph of arccot x and its derivative

Analyzing the differences of these functions through these graphs, you can observe that the original function $latex f(x) = \cot^{-1}{(x)}$ has a domain of

$latex (-\infty,\infty)$ or all real numbers

and exists within the range of

$latex (0,\pi)$ or $latex 0<y<\pi$

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

$latex (-\infty,\infty)$ or all real numbers

and exists within the range of

$latex [-1, 0)$ or $latex -1 \leq y < 0$


Examples

Below are some examples of using either the first or second method in deriving an inverse cotangent function.

EXAMPLE 1

Derive: $latex f(\theta) = \cot^{-1}{(\theta)}$

Solution: Analyzing the given inverse cotangent function, it is only an inverse cotangent of a single variable $latex \theta$ raised to a variable equal to one. Therefore, we can use the first method to derive this problem.

Step 1: Analyze if the inverse cotangent of $latex \theta$ is a function of $latex \theta$. In this problem, it is. Hence, proceed to step 2.

Step 2: Illustrate derivation through

$latex \frac{d}{d\theta} f(\theta) = \frac{d}{d\theta} \left(\cot^{-1}{(\theta)} \right)$

Step 3: Directly apply the derivative formula of inverse cotangent function and derive in terms of $latex \theta$. Since no further simplification is needed, the final answer is:

$latex f'(\theta) = -\frac{1}{1 + \theta^2}$

EXAMPLE 2

Derive: $latex F(x) = \cot^{-1}{\left(6x^2-3 \right)}$

Solution: Analyzing the given inverse cotangent function, it is an inverse cotangent of a polynomial function. Therefore, we can use the second method to derive this problem.

Step 1: Express the inverse cotangent function as $latex F(x) = \cot^{-1}{(u)}$ or $latex F(x) = \text{arccot}(u)$, where $latex u$ represents any function other than x. In this problem,

$latex u = 6x^2-3$

We will substitute this later as we finalize the derivative of the problem.

Step 2: Illustrate derivation by using the derivation symbols or denotations such as

$latex \frac{d}{dx} F(x) = \frac{d}{dx} \left(\cot^{-1}{(u)} \right)$

Step 3: Consider $latex \cot^{-1}{(u)}$ as the outside function $latex f(u)$ and $latex u$ as the inner function $latex g(x)$ of the composite function $latex F(x)$. For this problem, we have

$latex f(u) = \cot^{-1}{(u)}$

and

$latex g(x) = u = 6x^2-3$

Step 4: Get the derivative of the outer function $latex f(u)$, which must use the derivative of inverse cotangent $latex \cot^{-1}{(u)}$, in terms of $latex u$.

$latex \frac{d}{du} \left( \cot^{-1}{(u)} \right) = -\frac{1}{1+u^2}$

Step 5: Get the derivative of the inner function $latex g(x)$ or $latex u$. Since our $latex u$ in this problem is a polynomial function, we will use power rule and sum/difference of derivatives to derive $latex u$.

$latex \frac{d}{dx}(g(x)) = \frac{d}{dx} \left(6x^2-3 \right)$

$latex \frac{d}{dx}(g(x)) = 12x$

Step 6: Apply the basic chain rule formula by algebraically multiplying the derivative of outer function $latex f(u)$ by the derivative of inner function $latex g(x)$

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

$latex \frac{dy}{dx} = -\frac{1}{1+u^2} \cdot 12x$

Step 7: Substitute $latex u$ into $latex f'(u)$

$latex \frac{dy}{dx} = -\frac{1}{1+u^2} \cdot 12x$

$latex \frac{dy}{dx} = -\frac{1}{1+(6x^2-3)^2} \cdot 12x$

Step 8: Simplify and apply any function law whenever applicable to finalize the answer.

$latex F'(x) = -\frac{1}{1+(6x^2-3)^2} \cdot 12x$

$latex F'(x) = -\frac{12x}{1+(6x^2-3)^2}$

$latex F'(x) = -\frac{12x}{1 + 9(2x^2-1)^2}$

And the final answer is:

$latex F'(x) = -\frac{12x}{1 + 9(2x^2-1)^2}$

or

$latex F'(x) = -\frac{12x}{36x^4-36x^2+10}$


See also

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