# Derivative of arctan (Inverse Tangent) With Proof and Graphs

The Derivative of ArcTagent or Inverse Tangent is one of the commonly used transcendental functions in terms of getting their derivatives in Differential Calculus (or Calculus I). The derivative of the inverse tangent 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 arctangent or the inverse tangent function. We will cover brief fundamentals, its definition, formula, a graph comparison arctan and its derivative, a proof, methods to derive, and a few examples.

##### CALCULUS

Relevant for

Learning about the proof and graphs of the derivative or arctan of x.

See proof

##### CALCULUS

Relevant for

Learning about the proof and graphs of the derivative or arctan of x.

See proof

## Avoid confusion in using the denotations arctan(x), tan-1(x), 1 / tan(x) , and tann(x)

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

$latex \arctan{(x)} = \tan^{-1}{(x)}$

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

However, when it comes to the denotation $latex \tan^{-1}{(x)}$, sometimes it can confuse learners that $latex -1$ is an algebraic exponent of a non-inverse tangent, which is not true. The $latex -1$ used for inverse tangent represents the tangent being inverse and not raised to $latex -1$.

Therefore,

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

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

## Proof of the Derivative of the Inverse Tangent Function

In this proof, we will mainly use the concepts of a right triangle, the trigonometric function of tangent, 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, there is a side x perpendicular to the one-unit side and an angle y opposite to side x and adjacent to the one-unit side.

Using these components of a right-triangle, we can find the angle y by using Soh-Cah-Toa, particularly the tangent function since we have the adjacent and opposite sides of angle y.

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

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

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

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

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

$latex \frac{d}{dx} (\sec^{2}{(y)}) = 1$

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

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

We recall that based on trigonometric identities, $latex \sec^{2}{(\theta)} = 1 + \tan^{2}{(\theta)}$. Applying this, we have

$latex \frac{dy}{dx} = \frac{1}{1 + \tan^{2}{(y)}}$

From our given equation, we recall that

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

We can then substitude this to the implicitly derived equation.

$latex \frac{dy}{dx} = \frac{1}{1 + \tan^{2}{(y)}}$

$latex \frac{dy}{dx} = \frac{1}{1 + (\tan{(y)})^2}$

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

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

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

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

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

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

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

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

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

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

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

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

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

## How to derive an Inverse Tangent Function?

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

### METHOD 1: Derivative of the Inverse Tangent of any single variable x

Step 1: Analyze if the inverse tangent of a variable is a function of that same variable. For example, if the right-hand side of the equation is $latex \tan^{-1}{(x)}$, then check if it is a function of the same variable x or f(x). After this, proceed to Step 2 until you complete the derivation steps.

Note: If $latex \tan^{-1}{(x)}$ 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

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

or

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

Step 3: Then directly apply the derivative formula of inverse tangent 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 Tangent of any function u in terms of x.

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

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

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

or

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

Step 3: Consider $latex \tan^{-1}{(u)}$ 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

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

and

$latex g(x) = u$

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

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

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 Tangent x VS. The Derivative of Inverse Tangent x

Given the function

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

the graph is illustrated as

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

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

which is illustrated graphically as

Illustrating both graphs in one, we have

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

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

and exists within the range of

$latex \left( -\frac{\pi}{2},\frac{\pi}{2} \right)$ or $latex -\frac{\pi}{2}<y<\frac{\pi}{2}$

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

$latex (0, 1]$ or $latex 0 < y \leq 1$

## Examples

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

### EXAMPLE 1

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

Solution: Analyzing the given inverse tangent function, it is only an inverse tangent 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 tangent of $latex \theta$ is a function of $latex \theta$. In this problem, it is. Hence, proceed to step 2.

Step 2: Illustrate derivation through

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

Step 3: Directly apply the derivative formula of inverse tangent 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: $latex F(x) = \tan^{-1}{\left(3x^2+6 \right)}$

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

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

$latex u = 3x^2+6$

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

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

Step 3: Consider $latex \tan^{-1}{(u)}$ 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

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

and

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

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

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

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 $latex u$.

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

$latex \frac{d}{dx}(g(x)) = 6x$

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 $latex g(x)$

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

$latex \frac{dy}{dx} = \frac{1}{1+u^2} \cdot 6x$

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

$latex \frac{dy}{dx} = \frac{1}{1+u^2} \cdot 6x$

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

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

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

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

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

or

$latex F'(x) = \frac{6x}{9x^4+36x^2+37}$