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 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 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.
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
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
