Derivative of Arctan
- Introduction
- What Is the Derivative of Arctan?
- Proof of Derivative of Arctan by Chain Rule
- Proof of Derivative of Arctan by First Principle
- Real-life Application of Derivative of Arctan
- Solved Examples
- Practice Problems
- Frequently Asked Questions
Introduction
Arctan \(x\) is also known as the inverse of `tan` \(x\).
The derivative of the arctangent function or arctan derivative also is often denoted as \(\arctan(x)\) or \(\tan^{-1}(x)\), with respect to \(x\)
We know from composition of functions that if we take the composition of two functions that are inverses, then each will reverse the effect of the other. Meaning, if `f` and `f^-1` are inverse functions of each other then `f(f^-1(x)) = f^-1(f(x)) = x`. Likewise, `tan(arctan x) = arctan(tan x) = x`. We use these facts to find the derivative of arctan `x`.
What Is Derivative of Arctan?
The derivative of the arctangent function with respect to \(x\) is:
\( \frac{d}{dx}(\arctan(x)) = \frac{1}{1 + x^2} \)
In other words, if \(y = \arctan(x)\), then the derivative \(\frac{dy}{dx}\) is \(\frac{1}{1 + x^2}\). This result is derived using the chain rule in calculus.
It's worth noting that the derivative of the arctangent function is always positive, indicating that the function is strictly increasing. Additionally, the domain of the tan inverse function is all real numbers and the range is \((- \frac{\pi}{2}, \frac{\pi}{2})\).
Proof of Derivative of Arctan by Chain Rule:
To prove the derivative of \(\arctan(x)\) using the chain rule, let's start by defining the function \(y = \arctan(x)\). The arctangent function is defined as the inverse of the tangent function:
\( \tan(\arctan(x)) = x \)
Now, take the tangent of both sides:
\( \tan(y) = x \)
Now, differentiate both sides with respect to \(x\):
\( \sec^2(y) \frac{dy}{dx} = 1 \)
Solve for \(\frac{dy}{dx}\):
\( \frac{dy}{dx} = \frac{1}{\sec^2(y)} \)
Now, use the fact that \(\sec^2(y) = 1 + \tan^2(y)\) (from the Pythagorean identity for trigonometric functions):
\( \frac{dy}{dx} = \frac{1}{1 + \tan^2(y)} \)
Now, recall that \(\tan(y) = x\) from the definition of \(\arctan(x)\). Substitute this back in:
\( \frac{dy}{dx} = \frac{1}{1 + x^2} \)
So, we have shown that the derivative of \(\arctan(x)\) with respect to \(x\) is \(\frac{1}{1 + x^2}\) using the chain rule and the Pythagorean identity for trigonometric functions.
Proof of Derivative of Arctan by First Principle:
To find the derivative of \(\arctan(x)\) using the first principle,
Start with the expression of the first principle:
\( f'(x) = \lim_{{h \to 0}} \frac{\arctan(x + h) - \arctan(x)}{h} \)
Simplify by using the formula for the difference of arctangents,
\( \arctan(A) - \arctan(B) = \arctan\left(\frac{A-B}{1 + AB}\right) \),
Here, \( A=x+h \) and \( B= x \),
\( f'(x) = \lim_{{h \to 0}} \frac{\arctan\left(\frac{x + h - x}{1 + (x + h)x}\right)}{h} \)
\( = \lim_{{h \to 0}} \frac{\arctan\left(\frac{h}{1 + x^2 + hx}\right)}{h} \)
Expand the arctangent using its Maclaurin series expansion:
\( \arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots \)
\( = \lim_{{h \to 0}} \frac{1}{h} \left[ \frac{h}{1 + x^2 + hx} - \frac{h^3}{3(1 + x^2 + hx)^3} + \frac{h^5}{5(1 + x^2 + hx)^5} - \dots \right] \)
\( = \lim_{{h \to 0}} \left[ \frac{1}{1 + x^2 + hx} - \frac{h^2}{3(1 + x^2 + hx)^3} + \frac{h^4}{5(1 + x^2 + hx)^5} - \dots \right] \)
Now, applying the limit \(h \to 0\),
\( f'(x) = \frac{1}{1 + x^2 + 0} - 0 + 0 - \dots \)
\( = \frac{1}{1 + x^2} \)
It can be concluded that the derivative of \(\arctan(x)\) with respect to \(x\) is \( \frac{1}{1 + x^2} \).
Real-life Applications of Derivative of Arctan
The derivative of the arctangent function `(1/(1 + x^2))` has applications in various real-life scenarios, especially in fields related to physics, engineering, and signal processing. Here are a few examples:
`1`. Signal Processing: Control Systems and Robotics
In control systems and robotics, the arctangent function often appears in the context of calculating angles and orientations. The derivative of the arctangent function is useful in determining the rate of change of these angles, which is crucial for controlling the movement of robotic arms or stabilizing systems.
`2`. Physics: Angular Velocity
In physics, particularly in rotational motion, the derivative of the arctangent function is used to calculate angular velocity. For instance, in a system where an object's position depends on time and the angle of rotation, the derivative helps determine how fast the angle is changing.
`3`. Economics: Marginal Utility
In economics, the concept of marginal utility represents the additional satisfaction or benefit gained from consuming one more unit of a good. The derivative of the arctangent function may be involved in certain mathematical models that describe the relationship between consumption and utility.
`4`. Computer Graphics: `3D` Rendering
In computer graphics, especially in `3D` rendering, the arctangent function and its derivative play a role in determining angles and orientations of objects. This is crucial for realistic rendering, as it helps compute the lighting and shading effects on three-dimensional surfaces.
`5`. Navigation: Gyroscopes
In navigation systems, gyroscopes are used to measure orientation changes. The arctangent function and its derivative are involved in interpreting the gyroscope data to determine the angular displacement and rate of change, aiding in navigation and stabilization.
`6`. Communication Systems: Phase Detection
In communication systems, particularly in modulation schemes like phase modulation, the arctangent function and its derivative are used to detect and interpret phase changes in signals. This is vital in applications like wireless communication and digital signal processing.
Solved Examples
Example `1`: Find the derivative of \(f(x) = 2\arctan(3x)\).
Solution:
\(f(x) = 2\arctan(3x)\)
Apply the chain rule, where the derivative of \(\arctan(u)\) with respect to \(x\) is \(\frac{1}{1 + u^2} \cdot \frac{du}{dx}\):
\(\frac{df}{dx} = 2 \cdot \frac{1}{1 + (3x)^2} \cdot \frac{d}{dx}(3x)\)
\(= \frac{6}{1 + 9x^2}\)
So, the derivative of \(f(x) = 2\arctan(3x)\) is \(\frac{6}{1 + 9x^2}\).
Example `2`: Find the derivative of \(\arctan(2x)\) using the first principle.
Solution:
To find the derivative of \(\arctan(2x)\) using the first principle,
Start with the expression of the first principle:
\( f'(x) = \lim_{{h \to 0}} \frac{\arctan(2(x + h)) - \arctan(2x)}{h} \)
Simplify by using the formula for the difference of arctangents,
\( \arctan(A) - \arctan(B) = \arctan\left(\frac{A-B}{1 + AB}\right) \),
Here, \( A=2(x+h) \) and \( B= 2x \),
\( f'(x) = \lim_{{h \to 0}} \frac{\arctan\left(\frac{2x + 2h - 2x}{1 +2 (x + h)2x}\right)}{h} \)
\( = \lim_{{h \to 0}} \frac{\arctan\left(\frac{2h}{1 +4(x^2 + hx)}\right)}{h} \)
Expand the arctangent using its Maclaurin series expansion:
\( \arctan(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots \)
\( = \lim_{{h \to 0}} \frac{1}{h} \left[ \frac{2h}{1 +4( x^2 + hx)} - \frac{(2h)^3}{3(1 +4(x^2 + hx))^3} + \frac{(2h)^5}{5(1 +4(x^2 + hx))^5} - \dots \right] \)
\( = \lim_{{h \to 0}} \left[ \frac{2}{1 + 4(x^2 + hx)} - \frac{8h^2}{3(1 +4(x^2 + hx))^3} + \frac{32h^4}{5(1 + 4(x^2 + hx))^5} - \dots \right] \)
Now, applying the limit \(h \to 0\),
\( f'(x) = \frac{2}{1 +4 x^2 + 0} - 0 + 0 - \dots \)
\( = \frac{2}{1 +4 x^2} \)
The derivative of \(\arctan(2x)\) with respect to \(x\) is \( \frac{2}{1 + 4x^2} \).
Example `3`: Find the derivative of \(y = 5\arctan(2x)\).
Solution:
\( y = 5\arctan(2x) \)
Apply the chain rule:
\( \frac{dy}{dx} = 5 \cdot \frac{1}{1 + (2x)^2} \cdot \frac{d}{dx}(2x) \)
\( = \frac{10}{1 + 4x^2} \)
So, the derivative of \(y = 5\arctan(2x)\) is \(\frac{10}{1 + 4x^2}\).
Example `4`: Find the derivative of \(f(x) = \arctan(x^2 + 1)\).
Solution:
\( f(x) = \arctan(x^2 + 1) \)
Apply the chain rule:
\( \frac{df}{dx} = \frac{1}{1 + (x^2 + 1)^2} \cdot \frac{d}{dx}(x^2 + 1) \)
\( = \frac{2x}{1 + (x^2 + 1)^2} \)
So, the derivative of \(f(x) = \arctan(x^2 + 1)\) is \(\frac{2x}{1 + (x^2 + 1)^2}\).
Example `5`: Find the derivative of \(g(x) = \frac{\arctan(x)}{x}\).
Solution:
\( g(x) = \frac{\arctan(x)}{x} \)
Apply the quotient rule:
\( g'(x) = \frac{\frac{1}{1 + x^2} \cdot x - \arctan(x) \cdot 1}{x^2} \)
Simplify:
\( g'(x) = \frac{x - \arctan(x) - x^2 \arctan(x)}{x^2(1 + x^2)} \)
So, the derivative of \(g(x) = \frac{\arctan(x)}{x}\) is \(\frac{x - \arctan(x) - x^2 \arctan(x)}{x^2(1 + x^2)} \).
Practice Problems
Q`1`. Find the derivative of \(y = \arctan(3x)\).
- \(\frac{3}{1 + 9x^3}\)
- \(\frac{3}{1 + 9x^2}\)
- \(\frac{3}{1 + 9x^4}\)
- \(\frac{3}{2 + 9x^2}\)
Answer: b
Q`2`. Find the derivative of \(f(x) = \arctan(2x + 1)\).
- \(\frac{2}{1 + (2x + 1)^2}\)
- \(\frac{2}{3 + (2x + 1)^2}\)
- \(\frac{2}{1 + (2x + 1)^3}\)
- \(\frac{2}{1 + (3x + 1)^2}\)
Answer: a
Q`3`. Find the derivative of \(y = 4\arctan(5x)\).
- \(\frac{20}{1 + 20x^2}\)
- \(\frac{25}{1 + 25x^2}\)
- \(\frac{20}{1 + 25x^2}\)
- \(\frac{20}{1 + 25x^3}\)
Answer: c
Q`4`. Find the derivative of \(f(x) = \arctan(2x^2)\).
- \(\frac{2x}{1 + 4x^4}\)
- \(\frac{4x}{1 + 4x^2}\)
- \(\frac{4x}{1 + 2x^4}\)
- \(\frac{4x}{1 + 4x^4}\)
Answer: d
Q`5`. Find the derivative of \(f(x) = \arctan(e^{2x})\).
- \(\frac{2e^{2x}}{1 + e^{4x}}\)
- \(\frac{2e^{3x}}{1 + e^{4x}}\)
- \(\frac{2e^{2x}}{1 + e^{3x}}\)
- \(\frac{2e^{4x}}{1 + e^{4x}}\)
Answer: a
Frequently Asked Questions
Q`1`. What is the derivative of \(\arctan(x)\)?
Answer: The derivative of \(\arctan(x)\) with respect to \(x\) is \(\frac{1}{1 + x^2}\).
Q`2`. How is the chain rule applied to find the derivative of \(\arctan(3x)\)?
Answer: Applying the chain rule, the derivative of \(\arctan(3x)\) is \(\frac{3}{1 + 9x^2}\).
Q`3`. Can the quotient rule be used for \(\frac{d}{dx}(\arctan(x))\)?
Answer: No, the quotient rule is not directly applicable to \(\frac{d}{dx}(\arctan(x))\) as it's not a quotient of two functions.
Q`4`. Is the derivative of \(\arctan(x)\) always positive?
Answer: Yes, the derivative of \(\arctan(x)\) is always positive, indicating that \(\arctan(x)\) is a strictly increasing function.
Q`5`. What is the maximum value of the derivative of \(\arctan(x)\)?
Answer: \(\frac{d}{dx}(\arctan(x)) = \frac{1}{1 + x^2}\). It will be maximum when the denominator has the least value which is `1` in this case. Hence, the maximum value of the derivative of \(\arctan(x)\) is `1` at `x = 0`.