**The derivative of the inverse tangent function is equal to 1/(1+x ^{2}).** This derivative can be proved using the Pythagorean theorem and algebra.

In this article, we will discuss how to derive the arctangent or inverse tangent function. We’ll cover brief basics, a proof, a comparison graph of arctangent and its derivative, and some examples.

##### CALCULUS

**Relevant for**…

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

##### CALCULUS

**Relevant for**…

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

## Avoid confusion in using the denotations arctan(x), tan^{-1}(x),
1
/
tan(x) , and tan^{n}(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*.

## 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

The following are some examples of how to derive composite inverse tangent functions.

### EXAMPLE 1

Find the derivative of $latex f(x) = \tan^{-1}(4x)$

##### Solution

To derive this function, we can use the chain rule, since the inverse tangent function is composite.

If we consider $latex u=4x$ as the internal function, we have $latex f(u)=\tan^{-1}(u)$ and applying the chain rule, we have:

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

$$\frac{dy}{dx}=\frac{1}{1+u^2} \times 4$$

Then, we plug $latex u=4x$ back into the function and we have:

$$\frac{dy}{dx}=\frac{4}{1+(4x)^2}$$

$$\frac{dy}{dx}=\frac{4}{1+16x^2}$$

### EXAMPLE 2

Derive the function $latex F(x) = \tan^{-1}(x^2+2 )$

##### Solution

We can write the inverse tangent function as $latex f (u) = \tan^{-1}(u)$, where $latex u = x^2+2$.

Therefore, to use the chain rule, we start by finding the derivative of the outer function $latex f(u)$:

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

Then, we obtain the derivative of the inner function $latex g(x)=u=x^2+2$:

$$\frac{d}{dx}(g(x)) = \frac{d}{dx}(x^2+2)$$

$$\frac{d}{dx}(g(x)) = 2x$$

By the chain rule, we multiply the derivative of the external function by the derivative of the internal function and we have:

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

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

Finally, we substitute $latex u$ back in and simplify:

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

$$\frac{dy}{dx} = \frac{2x}{1+x^4+4x^2+4}$$

$$F'(x) = \frac{2x}{x^4+4x^2+5}$$

### EXAMPLE 3

What is the derivative of $latex f(x) = \tan^{-1}(\sqrt{x})$?

##### Solution

To derive this function, we consider the square root as the inner function. Then, we write $latex u=\sqrt{x}$ as $latex u=x^{\frac{1}{2}}$, to find the following derivative:

$$\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}$$

Now, let’s write $latex f(u)=\tan^{-1}(u)$ and apply the chain rule:

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

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

Substituting $latex u=\sqrt{x}$ and simplifying, we have:

$$\frac{dy}{dx}=\frac{1}{1+(\sqrt{x})^2} \times \frac{1}{2}x^{-\frac{1}{2}}$$

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

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

## Practice of derivatives of composite inverse tangent functions

## See also

Interested in learning more about the derivatives of trigonometric functions? Take a look at these pages:

- Derivative of arcsin (Inverse Sine) With Proof and Graphs
- Derivative of arccos (Inverse Cosine) With Proof and Graphs
- Derivative of arcsec (Inverse Secant) With Proof and Graphs
- Derivative of arccsc (Inverse Cosecant) With Proof and Graphs
- Derivative of arccot (Inverse Cotangent) With Proof and Graphs

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