Last updated on July 4th, 2025
Derivative of arctan(x) is 1/1 + x², shows how the slope of the arctan(x) curve changes depending on the value of x. This is used in navigation, robotics, signal processing, and projectile trajectories. In this topic, we will learn more about the derivative of arctan.
Trigonometric differentiation is the process of determining the derivatives of trigonometric functions. The derivative of arctan x is represented by d/dx (arctan x) or d/dx (tan-1x) or (arctan x) or (tan-1x).
This is the inverse function to find the derivative of the inverse of the tangent function: y = tan-1x = arctan x. The rate of change of the arc tangent function with respect to the independent variable is given by its derivative.
The formula to find the derivative of the inverse tangent function or simply the arctan is given below:
d/dx (arctan x) = 1/(1 + x2)
Its proofs can be explored through different approaches: using the definition of a derivative (first principle rule), applying the chain rule by relating it to the tangent function, and leveraging the product rule with carefully chosen expressions. Here is the detailed method of using all the three proofs:
To derive the derivative of arctan (x) using the first principle (also known as the definition of the derivative), we follow these steps:
Step 1: The derivative of a function f (x) at a point x = a is defined as:
f′(a) = h→0 f (a + h) - f (a)/h
For the function f (x) = arctan (x), we want to find f′(x), so we apply the definition at x rather than a specific point a:
d/dx arctan (x) = h→0 arctan (x + h) - arctan (x)/h
Step 2: The difference of two arc tangents can be simplified using the following identity:
arctan (a) – arctan (b) = arctan a - b/1 + ab
Apply this identity to arctan (x + h) – arctan (x):
Here, a is (x + h) and b is x.
arctan (x + h) – arctan (x)
= arctan (x + h) - x/1 + (x + h) x
By simplifying the equation x + h - x, the x and -x get canceled and the result becomes h.
= arctan h/1 + x (x + h)
Step 3: Substitute this expression into the original derivative formula
d/dx arctan (x) = h→0 arctan h/1 + x (x + h)/h
Step 4: Simplify the limit
As h → 0, the expression inside the arctangent becomes:
h/1 + x2 + xh
For very small h, the term xh becomes negligible, so this simplifies to:
h/1 + x2
Thus, the expression becomes
d/dx arctan (x) = h→0 arctan h/ 1 + x2/h
Since arctan (y) ≈ y for small value of y, we can approximate
arctan (h/1 + x2) as h/1 + x2.
This leads to:
d/dx arctan (x) = 1/1 + x2
Thus, the derivative of arctan x is:
d/dx arctan (x) = 1/1 + x2
To find the formula or the proof of this formula, use the formula for the derivative of an inverse function to find the derivative of the inverse function of the tangent function. That is: y = tan-1 x = arctan x.
Step 1: Simplify the equation by taking the tangent of both sides
y = tan-1x
tan y = tan (tan-1x)
tan y = x
Step 2: You may know that:
d/dy tan y = d/dy sin y/cos y
= 1/cos2 y
= sec2 y
Step 3: Now, use the implicit differentiation to take the derivative of both sides of our original equation to get:
tan y = x
d/dx (tan (y)) = d/dx x
Step 4: By following the Chain Rule
d/dy tan (y) dy/dx = 1
(1/cos2 (y)) dy/dx = 1
dy/dx = cos2(y)
Step 5: Using the Pythagorean theorem, that is hypotenuse is:
(Hypotenuse)2 = (Opposite)2 + (Adjacent)2
By substituting the values:
(Hypotenuse)2 = x2 + 12
To solve for the hypotenuse, we should take the square root on both sides
h = √1 + x2
Step 6: We can apply the formula and compute,
cos(y) = 11 + x2
Step 7: From this, we get:
cos2(y) = 11 + x22 = 11 + x2
So,
dydx = 11 + x2
In other words,
ddx arctan (x) = 11 + x2
The derivative of arctan(x) can be derived using the product rule by expressing it as a product of functions that simplify during differentiation. Let’s understand the proof in detail:
Step 1: Start by recognizing that arctan (x) can be written as an inverse function, but to use the product rule, we can manipulate it in terms of a product. We use the identity:
arctan (x) = 12i ln 1 + ix1 - ix
Step 2: To differentiate this expression, recall the product rule, which states:
ddx u (x) v (x) = u′(x)v(x)+u(x)v′(x)
Step 3: Differentiate using the product rule
We can differentiate this expression using the product rule. The function involves a constant i2i and the logarithm function ln 1 + ix1 - ix.
The derivative of a constant times a function is simply the constant times the derivative of the function, so we focus on differentiating the logarithmic part:
ddx 12i ln 1 + ix1 - ix
Step 4: Use the quotient rule to differentiate the argument 1 + ix1 - ix
ddx 1 + ix1 - ix
= (1 - ix) i - (1 + ix) (-i) (1 - ix)2
Simplify the numerator:
(1 – ix) i – (1 + ix) (-i)
= i – i2x + i + ix
= 2i + 2x
Thus,
2i + 2x(1 - ix)2
Step 5: Now, combine everything. The derivative of the logarithmic function is:
ddx ln 1 + ix1 - ix = 2i + 2x(1 - ix)2
Now, simplify this expression by multiplying the terms and simplifying the complex fractions. After simplifying, the result reduces to:
11 + x2
Step 6: Simplify using properties of the logarithm
ddx arctan (x) = 11 + x2 11 + ix1 - ix
ddx arctan (x) = 11 + x2
The higher-order derivatives arctan (x) involve finding the derivatives of its first derivative and continuing to differentiate successively to obtain more complex expressions. These higher derivatives provide insights into the behavior of the arctangent function and are useful in various applications such as series expansions and approximations.
Nth Derivative of Arctan
The n-th derivative of arctan (x) can be found by differentiating repeatedly. The general formula for the n-th derivative of arctan (x) is given by:
dndxn arctan (x) = (-1)n-1 (n - 1)!(1 + x2)n
Special Cases
The first derivative of arctan (x) is 11 + x2. At x = 0, this simplifies to:
ddx arctan (x)x=0 = 11 + 02 = 1
So, the slope of the tangent line to the curve y = arctan (x) at x = 0 is 1.
As x
As x approaches infinity, the derivative of arctan (x) approaches zero:
x11 + x2 = 0
This means that as x becomes very large, the slope of the curve y = arctan (x) becomes almost horizontal.
As x –
Similarly, as x approaches negative infinity, the derivative of arctan (x) also approaches zero:
x- 11 + x2 = 0
Thus, the slope of the curve approaches zero as x becomes very negative.
At x = 1
At x = 1, the derivative of arctan (x) is:
ddx arctan (x)x = 1 = 11 + 12 = 12
This means that the slope of the tangent line to the curve at x = 1 is 12.
At x = – 1
At x = –1, the derivative of arctan (x) is:
ddx arctan (x)x = -1 = 11 + (-1)2 = 12
This means that the slope of the tangent line to the curve at x = –1 is also 1/2.
There are certain mistakes students might make while doing the process of the derivative of arctan. Here are the five mistakes and how to solve them:
Find the derivative of y = arctan (x)
The derivative of arctan (x) is d/dx = 1/1 + x2
This is the direct application of the formula for the derivative of arctan (x). It tells us how arctan (x) changes with respect to x.
Differentiate y = arctan (3x)
d/dx = 3/1 + 9x2
The chain rule is applied because y = arctan (u), where u = 3x. First, we find the derivative of arctan (u) as 1/1 + u2
then multiply it by the derivative of u = 3x.
If arctan (y) = x², find dy/dx.
dy/dx = (1 + y2) 2x
Implicit differentiation is used since y is a function of x. The chain rule is applied to arctan (y), followed by solving for dy/dx.