*This post may contain affiliate links. If you make a purchase through links on our site, we may earn a commission.*

In this article, we’ll focus on understanding the process of determining the derivative of arctan and use it to simplify the derivative of other functions. Knowing how to find the inverse tangent’s derivative expression will also help us integrate expressions once we’re taking integral calculus lessons.

**The derivative of arctan returns an algebraic expression. We can use this expression to find the derivative of inverse trigonometric functions.**

In our discussion, we’ll understand how to differentiate arctan and use the new derivative rule to differentiate more complex functions. Make sure to have your notes handy!

**What is the derivative of arctan?**

The derivative of arctan or $y = \tan^{-1}x$ can be determined using the formula shown below.

\begin{aligned}\tan y = x &\Leftrightarrow y = \tan^{-1} x\\\dfrac{d}{dx} \tan^{-1} x &= \dfrac{1}{1 + x^2} \end{aligned}

Recall that the inverse tangent of $x$ is simply the value of the angle, $y$ in radians, where $\tan y = x$. This shows that the derivative of the inverse tangent function is indeed an algebraic expression. This means that if $f(x) = \tan^{-1} x$, $f’(a)$ is simply equal to the reciprocal of $1 + a^2$.

For example, if we want to find $f’\left(\dfrac{1}{\sqrt{3}}\right)$, we can simply substitute $x = \dfrac{1}{\sqrt{3}}$ into the formula for the derivative of arctan, $f’(x) = \dfrac{1}{1 + x^2}$.

\begin{aligned}f(x)&= \tan^{-x}\\f'(x)&= \dfrac{1}{1 +x^2}\\\\f’\left({\color{Teal}\dfrac{1}{\sqrt{3}}} \right )&= \dfrac{1}{1 +\left({\color{Teal}\dfrac{1}{\sqrt{3}}} \right )^2}\\&= \dfrac{1}{1 + \dfrac{1}{3}} \\&= \dfrac{3}{4}\end{aligned}

We can also use the derivative rule for arctan to differentiate functions that contain tangent inverse (or arctan) in its expression. Before we do so, let’s take a quick look on how we were able to come up with the derivative rule, $\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1+x^2}$.

**How to derive arctan within a function?**

As we have mentioned, when $y = \tan^{-1} x$, this means that $\tan y = x$. We can then take the derivative of both sides of the new equation. Keep in mind that we’ll have to use the chain rule for $\tan y$ as shown below.

**Derivative of arctan proof**

\begin{aligned}y&= \tan^{-1}x\\\tan y &= x\\\dfrac{d}{dx}\tan y &= \dfrac{d}{dx}x\\ {\color{Teal}\sec^2 y} \cdot {\color{Purple}\dfrac{dy}{dx}} &= \dfrac{d}{dx}x,\phantom{x}{\color{Teal}\text{Derivative of Tangent }}\text{&}{\color{Purple}\text{ Chain Rule }}\\\dfrac{d}{dx}\tan y &= \dfrac{d}{dx}x\\ {\sec^2 y} \cdot {\dfrac{dy}{dx}} &= {\color{Teal}1},\phantom{x}{\color{Teal}\text{Power Rule }}\end{aligned}

We can divide both sides by $\sec^2 y$ and rewrite $\dfrac{1}{\cos^ y}$ as $\sec^2 y$.

\begin{aligned} {\sec^2 y} \cdot {\dfrac{dy}{dx}} &= 1\\\dfrac{dy}{dx} &= \dfrac{1}{\sec^2 y}\\\dfrac{dy}{dx} &= \cos^2 y\\y’ &= \cos^2 y\end{aligned}

Remember that $y = \tan^{-1} x$, so we can use this expression to rewrite $y’$.

\begin{aligned} y’ &= \cos^2 ( \tan^{-1} x)\end{aligned}

We can use the right triangle shown below to show the fact that $\cos^2 (\tan^{-1} x)$ is indeed equal to $\dfrac{1}{1 + x^2}$.

By visually representing $\tan y = x$ and $y =\tan^{-1} x$ on a right triangle and using the Pythagorean theorem, $c^2 =a^2 +b^2$, we can come up with the following.

\begin{aligned}\cos y &= \dfrac{1}{\sqrt{1 +x^2}}\\ \cos^2 y &= \dfrac{1}{1 + x^2}\end{aligned}

We can substitute this expression into our current expression for $y’$ as shown below.

\begin{aligned} y’ &= \cos^2 ( \tan^{-1} x)\\&= \dfrac{1}{1 + x^2}\end{aligned}

This confirms our derivative rule, $\dfrac{d}{dx} \tan^{-1} x = \dfrac{1}{1 + x^2}$.

Now that we’ve proven the derivative of arctan, it’s time that we use this to differentiate functions that contain $\tan^{-1}x$ within its terms or contain $\tan^{-1} g(x)$. When you’re ready, try out the examples we’ve prepared for you.

**Example 1**

Find the derivative of $h(x) = \tan^{-1} \sqrt{x}$.

__Solution__

As with differentiating other inverse trigonometric functions, it’s helpful to remember that the chain rule will apply if we have another function (other than $x$, of course) inside $\tan^{-1} x$. We can rewrite $h(x)$ as $f[g(x)]$, where $f(x)$ is the outer function and $g(x)$ is the inner function.

\begin{aligned}h’(x) &= \dfrac{d}{dx} f[g(x)] \\&= f'[g(x)] \cdot g'(x)\\f(x) &= \tan^{-1} x\\ g(x) &= \sqrt{x}\\f[g(x)] &= \tan^{-1} (\sqrt{x})\end{aligned}

We can now use the formula for the derivative of arctan, $\dfrac{d}{dx}\tan^{-1}x =\dfrac{1}{1+x^2}$.

\begin{aligned}h'(x)&= {\color{Teal}\dfrac{1}{1 + [g(x)]^2}} \cdot g'(x),\phantom{x} {\color{Teal} \text{Derivative of Tangent Inverse}}\\&= \dfrac{1}{1 + (\sqrt{x})^2} \cdot \dfrac{d}{dx} \sqrt{x}\\&= \dfrac{1}{1 + (\sqrt{x})^2} \cdot \dfrac{d}{dx} x^{\frac{1}{2}}\\&= \dfrac{1}{1 + x} \cdot {\color{Teal}\dfrac{1}{2}x^{\frac{1}{2} -1}},\phantom{x} {\color{Teal} \text{Power Rule}}\\&= \dfrac{1}{2(1 + x)}\cdot x^{-\frac{1}{2}}\\&= \dfrac{1}{2\sqrt{x}(1 + x)}\end{aligned}

This shows that it is important for us to master our fundamental derivative rules even when we’re finding the derivatives of inverse trigonometric functions or functions that contain $\tan^{-1} x$. Hence, we have $h’(x) = \dfrac{1}{2\sqrt{x}(1 + x)}$.

**Example ****2**

Find the derivative of $h(x) = \tan^{-1} [e^x(x^3 + 3x)]$.

__Solution__

We will be using the chain rule to differentiate $h(x)$, so we will need the derivative of $e^x(x^3 + 3x)$. We can do this first by using the product rule, $\dfrac{d}{dx} [f(x)g(x)] = f’(x) g(x) + g’(x)f(x)$, where $f(x)= e^x$ and $g(x) = x^3 + 3x$ for our case.

\begin{aligned}\dfrac{d}{dx}[e^x(x^3 + 3x)] &= e^x \dfrac{d}{dx}(x^3 + 3x) + (x^3 + 3x)\dfrac{d}{dx} e^x \\&= e^x \dfrac{d}{dx}(x^3 + 3x) + (x^3 + 3x){\color{Teal} e^x},\phantom{x}{\color{Teal}\text{Derivative of }e^x} \\&= e^x {\color{Teal}\left(\dfrac{d}{dx} x^3 + \dfrac{d}{dx}3x\right)} + (x^3 + 3x)e^x,\phantom{x}{\color{Teal}\text{Sum Rule}}\\&= e^x (\dfrac{d}{dx} x^3+ {\color{Teal}3\dfrac{d}{dx} x })+ (x^3 + 3x)e^x,\phantom{x}{\color{Teal}\text{Constant Multiple Rule}}\\&= e^x ({\color{Teal}3x^2 + 3 })+ (x^3 + 3x)e^x,\phantom{x}{\color{Teal}\text{Power Rule}}\\&= e^x(x^3+3x^2+3x+3) \end{aligned}

Let’s now focus on our actual function, $h(x) = \tan^{-1} [e^x(x^3 + 3x)]$, and apply the chain rule as shown below.

\begin{aligned}h’(x) &= \dfrac{d}{dx} f[g(x)] \\&= f'[g(x)] \cdot g'(x)\\f(x) &= \tan^{-1} x\\ g(x) &= e^x(x^3 + 3x)\\f[g(x)] &= \tan^{-1}[e^x(x^3 + 3x)]\end{aligned}

\begin{aligned}h'(x)&= {\color{Teal}\dfrac{1}{1 + [g(x)]^2}} \cdot g'(x),\phantom{x} {\color{Teal} \text{Derivative of Tangent Inverse}}\\&= \dfrac{1}{1 + [e^x(x^3 + 3x)]^2} \cdot \dfrac{d}{dx} [e^x(x^3 + 3x)]\\&= \dfrac{1}{1 + e^{2x}(x^3 + 3x)^2} \cdot \dfrac{d}{dx} [e^x(x^3 + 3x)]\end{aligned}

What we can do now is replace $\dfrac{d}{dx} [e^x(x^3 + 3x)]$ with its derivative, $e^x(x^3+3x^2+3x+3)$.

\begin{aligned}h'(x)&= \dfrac{1}{1 + e^{2x}(x^3 + 3x)^2} \cdot {\color{Teal}e^x(x^3 + 3x^2 + 3x + 3)}, \phantom{x}{\color{Teal} \text{Product Rule}}\\&= \dfrac{e^x(x^3 + 3x^2 + 3x + 3)}{1 + e^{2x}(x^3 + 3x)^2}\end{aligned}

Hence, through fundamental derivative rules and the derivative of arctan, we have $h’(x) = \dfrac{e^x(x^3 + 3x^2 + 3x + 3)}{1 + e^{2x}(x^3 + 3x)^2}$.

**Practice Questions**

1. Find the derivative of the following functions.

a. $f(x) = \tan^{-1} (2x^2 -1)$

b. $g(x) = \tan^{-1} \sqrt[3]{x}$

c. $h(x) = \tan^{-1} \left(\dfrac{1}{2x + 1}\right)$

2. Find the derivative of the following functions.

a. $f(x) = \tan^{-1} \sqrt{4x^2 + 5}$

b. $g(x) = \tan^{-1} \left(\dfrac{\ln x}{4x – 1}\right)$

c. $h(x) = \tan^{-1} [e^x(x^4 – 4x)]$

**Answer Key**

1.

a. $f’(x) = \dfrac{2x}{2x^4-2x^2+1}$

b. $g’(x) =\dfrac{1}{3x^{\frac{2}{3}}\left(x^{\frac{2}{3}}+1\right)}$

c. $h’(x) =-\dfrac{1}{2x^2+2x+1}$

2.

a. $f’(x) = \dfrac{2x}{(2x^2 +3)\sqrt{4x^2 + 5}}$

b. $g’(x) =\dfrac{4x-1-4x\ln \left(x\right)}{x \left(\ln ^2\left(x\right) + \left(4x- 1\right)^2\right)}$

c. $h’(x) =\dfrac{e^x\left(x^4 – 4x\right) + e^x\left(4x^3 – 4\right)}{e^{2x}\left(x^4 – 4x\right)^2 +1}$

*Images/mathematical drawings are created with GeoGebra.*

As an expert and enthusiast, I have a vast amount of knowledge on various topics, including calculus and mathematical concepts. I can provide information and insights on the process of determining the derivative of arctan and its applications in simplifying the derivative of other functions.

To begin, let's understand the derivative of arctan. The derivative of arctan, or y = tan^(-1)(x), can be determined using the formula:

d/dx(tan^(-1)(x)) = 1 / (1 + x^2) This formula shows that the derivative of the inverse tangent function is an algebraic expression. If f(x) = tan^(-1)(x), then f'(a) is equal to the reciprocal of (1 + a^2).

For example, if we want to find f'(1/sqrt(3)), we can substitute x = 1/sqrt(3) into the formula for the derivative of arctan:

f(x) = tan^(-1)(x) f'(x) = 1 / (1 + x^2)

f'(1/sqrt(3)) = 1 / (1 + (1/sqrt(3))^2) = 1 / (1 + 1/3) = 3/4 We can also use the derivative rule for arctan to differentiate functions that contain the tangent inverse (or arctan) in their expressions. The process involves using the chain rule and the derivative of arctan to differentiate more complex functions.

To derive arctan within a function, we start with the equation y = tan^(-1)(x), which implies tan(y) = x. We then take the derivative of both sides of the equation. Using the chain rule for tan(y), we have:

d/dx(tan(y)) = d/dx(x) (sec^2(y)) * (dy/dx) = 1 Dividing both sides by sec^2(y) and rewriting 1/cos^2(y) as sec^2(y), we get:

(dy/dx) = cos^2(y) Since y = tan^(-1)(x), we can substitute this expression into the derivative:

(dy/dx) = cos^2(tan^(-1)(x)) Using a right triangle representation and the Pythagorean theorem, we can show that cos^2(tan^(-1)(x)) is equal to 1 / (1 + x^2):

cos(tan^(-1)(x)) = 1 / sqrt(1 + x^2) cos^2(tan^(-1)(x)) = 1 / (1 + x^2) Therefore, the derivative rule for arctan is confirmed: d/dx(tan^(-1)(x)) = 1 / (1 + x^2).

Now, let's apply these concepts to the given examples:

Example 1: Find the derivative of h(x) = tan^(-1)(sqrt(x)).

To differentiate h(x), we can use the chain rule. Let f(x) = tan^(-1)(x) and g(x) = sqrt(x). Then, h(x) can be written as f(g(x)):

h'(x) = f'(g(x)) * g'(x)

Using the derivative rule for arctan, f'(x) = 1 / (1 + x^2), and the derivative of sqrt(x), g'(x) = 1 / (2 * sqrt(x)), we can calculate h'(x):

h'(x) = (1 / (1 + (sqrt(x))^2)) * (1 / (2 * sqrt(x)))
= 1 / (2 * sqrt(x) * (1 + x))

Therefore, h'(x) = 1 / (2 * sqrt(x) * (1 + x)).

Example 2: Find the derivative of h(x) = tan^(-1)(e^x(x^3 + 3x)).

Again, we can use the chain rule to differentiate h(x). Let f(x) = tan^(-1)(x) and g(x) = e^x(x^3 + 3x). Then, h(x) can be written as f(g(x)):

h'(x) = f'(g(x)) * g'(x)

Using the derivative rule for arctan, f'(x) = 1 / (1 + x^2), and the product rule to differentiate g(x), we can calculate h'(x):

h'(x) = (1 / (1 + (e^x(x^3 + 3x))^2)) * (e^x(x^3 + 3x))'

To find (e^x(x^3 + 3x))', we can apply the product rule:

(e^x(x^3 + 3x))' = (e^x)' * (x^3 + 3x) + e^x * (x^3 + 3x)'

The derivative of e^x is e^x, and the derivative of (x^3 + 3x) is 3x^2 + 3. Therefore:

(e^x(x^3 + 3x))' = e^x * (x^3 + 3x) + e^x * (3x^2 + 3)
= e^x(x^3 + 3x + 3x^2 + 3)

Substituting this back into h'(x), we have:

h'(x) = (1 / (1 + (e^x(x^3 + 3x))^2)) * (e^x(x^3 + 3x + 3))

Therefore, h'(x) = (e^x(x^3 + 3x + 3)) / (1 + (e^x(x^3 + 3x))^2).

I hope this explanation helps you understand the process of determining the derivative of arctan and its applications in differentiating functions. If you have any further questions, feel free to ask!