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Last updated on September 27, 2025
We use the derivative of tan⁻¹(x), which is 1/(1 + x²), as a tool to understand how the inverse tangent function changes with respect to slight changes in x. Derivatives are essential for calculating rates of change in various real-life situations. We will now discuss the derivative of tan⁻¹(x) in detail.
We now understand the derivative of tan⁻¹x. It is commonly represented as d/dx (tan⁻¹x) or (tan⁻¹x)', and its value is 1/(1 + x²). The function tan⁻¹x has a clearly defined derivative, indicating it is differentiable for all real numbers.
The key concepts are mentioned below: Inverse Tangent Function: (tan⁻¹(x) is the inverse of tan(x)). Chain Rule: A rule for differentiating composite functions. Secant Function: sec(x) = 1/cos(x).
The derivative of tan⁻¹x can be denoted as d/dx (tan⁻¹x) or (tan⁻¹x)'. The formula we use to differentiate tan⁻¹x is: d/dx (tan⁻¹x) = 1/(1 + x²)
The formula applies to all real x.
We can derive the derivative of tan⁻¹x using proofs. To show this, we will use implicit differentiation and trigonometric identities.
There are several methods to prove this, such as:
We will now demonstrate that the differentiation of tan⁻¹x results in 1/(1 + x²) using these methods:
Consider y = tan⁻¹x. This implies x = tan(y). Differentiating both sides with respect to x gives: 1 = sec²(y) · dy/dx Recall that sec²(y) = 1 + tan²(y). Thus, 1 = (1 + x²) dy/dx (since tan(y) = x) Therefore, dy/dx = 1/(1 + x²).
We know the identity tan²(y) + 1 = sec²(y). Therefore, using implicit differentiation, we substitute tan(y) = x, resulting in: dy/dx = 1/(1 + x²).
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a bit complex. To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives help analyze functions like tan⁻¹(x).
For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.
For the nth Derivative of tan⁻¹(x), we generally use fⁿ(x) for the nth derivative of a function f(x), which tells us the change in the rate of change (continuing for higher-order derivatives).
When x = 0, the derivative of tan⁻¹x = 1/(1 + 0²), which is 1.
Students frequently make mistakes when differentiating tan⁻¹x. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of tan⁻¹(3x).
Let y = tan⁻¹(3x). Using the chain rule, dy/dx = d/dx [tan⁻¹(3x)] = 3/(1 + (3x)²) = 3/(1 + 9x²).
We find the derivative by recognizing tan⁻¹(3x) as a composite function.
We apply the chain rule to differentiate the outer function tan⁻¹ and then the inner function 3x, combining the results.
The height of a water tank is given by the function h(x) = tan⁻¹(x), where x is the time in hours. Find the rate of change of the height when x = 1 hour.
Given h(x) = tan⁻¹(x), the rate of change is the derivative: dh/dx = 1/(1 + x²).
Substitute x = 1: dh/dx = 1/(1 + 1²) = 1/2.
Thus, the rate of change of the height at x = 1 hour is 1/2 units per hour.
To find the rate of change, we differentiate the function h(x) = tan⁻¹(x) and evaluate it at x = 1. The result indicates how the height changes with respect to time at that specific moment.
Derive the second derivative of the function y = tan⁻¹(x).
First, find the first derivative: dy/dx = 1/(1 + x²).Now differentiate again to find the second derivative: d²y/dx² = d/dx [1/(1 + x²)] = -2x/(1 + x²)². Therefore, the second derivative of the function y = tan⁻¹(x) is -2x/(1 + x²)².
We use implicit differentiation to find the first derivative, then differentiate once more, applying the quotient rule to obtain the second derivative.
Simplifying gives the final answer.
Prove: d/dx (tan⁻¹(x²)) = 2x/(1 + x⁴).
Consider y = tan⁻¹(x²). Using the chain rule: dy/dx = d/dx [tan⁻¹(x²)] = 1/(1 + (x²)²) · d/dx (x²) = 1/(1 + x⁴) · 2x = 2x/(1 + x⁴). Hence proved.
In this process, we applied the chain rule to differentiate tan⁻¹(x²).
We first differentiate the outer function tan⁻¹ and then multiply by the derivative of the inner function x², resulting in the final expression.
Solve: d/dx (tan⁻¹(x²)/x).
To differentiate the function, we use the quotient rule: d/dx (tan⁻¹(x²)/x) = (x · d/dx (tan⁻¹(x²)) - tan⁻¹(x²) · d/dx (x))/x². Substitute d/dx (tan⁻¹(x²)) = 2x/(1 + x⁴): = (x · (2x/(1 + x⁴)) - tan⁻¹(x²) · 1)/x² = (2x²/(1 + x⁴) - tan⁻¹(x²))/x². Therefore, d/dx (tan⁻¹(x²)/x) = (2x²/(1 + x⁴) - tan⁻¹(x²))/x².
In this process, we differentiate the given function using the quotient rule.
We apply the derivative of tan⁻¹(x²) and simplify the equation to obtain the final result.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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