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Last updated on September 9, 2025
The derivative of tan⁻¹(x), often denoted as d/dx (tan⁻¹x) or (tan⁻¹x)', is a crucial tool in calculus for understanding how the inverse tangent function changes in response to variations in x. Derivatives play a significant role in various real-life applications such as physics and engineering. In this discussion, we will explore the derivative of tan⁻¹x in detail.
The derivative of tan⁻¹x is represented as d/dx (tan⁻¹x) or (tan⁻¹x)'. The value of this derivative is 1/(1+x²).
The inverse tangent function has a well-defined derivative, indicating it is differentiable across its domain.
Key concepts include:
Inverse Tangent Function: (tan⁻¹x is the inverse of the tangent function).
Chain Rule: Used for differentiating composite functions involving tan⁻¹x.
Domain Consideration: The function is differentiable for all real x.
The derivative of tan⁻¹x is expressed as: d/dx (tan⁻¹x) = 1/(1+x²) This formula is applicable for all real numbers x.
We can derive the derivative of tan⁻¹x using several proofs, employing trigonometric identities and differentiation rules. Here are some methods:
By First Principle
Using Chain Rule
We will demonstrate that the derivative of tan⁻¹x results in 1/(1+x²) using these methods:
By First Principle
The derivative of tan⁻¹x can be derived using the First Principle, expressing it as the limit of the difference quotient.
Let f(x) = tan⁻¹x. Its derivative is: f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given f(x) = tan⁻¹x, write f(x + h) = tan⁻¹(x + h).
Substitute into the limit: f'(x) = limₕ→₀ [tan⁻¹(x + h) - tan⁻¹x] / h
Using the identity for tan⁻¹(a) - tan⁻¹(b), f'(x) = limₕ→₀ [1/(1+(x+h)x)(h)] = limₕ→₀ 1/(1+x²)
Hence, f'(x) = 1/(1+x²).
Using Chain Rule
To prove the differentiation of tan⁻¹x using the chain rule,
Consider y = tan⁻¹x.
Express tan⁻¹x = y then x = tan(y).
Differentiating both sides, dx/dy = sec²(y) Using the identity sec²(y) = 1 + tan²(y), dx/dy = 1 + x² Thus, dy/dx = 1/(1 + x²).
Therefore, the derivative of tan⁻¹x is 1/(1+x²).
Higher-order derivatives involve differentiating a function multiple times.
For example, the speed of a car changes (first derivative), and the rate at which that speed changes (second derivative).
Higher-order derivatives of tan⁻¹x provide deeper insights into its behavior.
First Derivative: f′(x) = 1/(1+x²).
Second Derivative: The derivative of f′(x) provides more information about the curvature.
Nth Derivative: fⁿ(x) represents the nth derivative of the function, indicating changes in the rate of change.
When x = 0, the derivative of tan⁻¹x = 1/(1+0²), which is 1. As x approaches ±∞, the derivative approaches 0, indicating the function levels off.
Students often make errors when differentiating tan⁻¹x. Understanding the correct approach can resolve these mistakes. Here are some common errors and solutions:
Calculate the derivative of tan⁻¹(x)·(1+x²).
Let f(x) = tan⁻¹(x)·(1+x²).
Using the product rule, f'(x) = u′v + uv′
Here, u = tan⁻¹x and v = 1+x².
Differentiate each term: u′ = d/dx (tan⁻¹x) = 1/(1+x²) v′ = d/dx (1+x²) = 2x
Substitute back into the equation: f'(x) = [1/(1+x²)](1+x²) + (tan⁻¹x)(2x) = 1 + 2x tan⁻¹x
Thus, the derivative is 1 + 2x tan⁻¹x.
We find the derivative by breaking the function into two parts, differentiating each, and then applying the product rule to combine them for the final result.
A bridge's slope is modeled by y = tan⁻¹(x). If x = 1 meter, find the slope of the bridge.
Given y = tan⁻¹(x), the slope is represented by dy/dx.
Differentiate y = tan⁻¹(x): dy/dx = 1/(1+x²)
Substitute x = 1: dy/dx = 1/(1+1²) = 1/2
Hence, the slope of the bridge at x = 1 meter is 1/2.
We compute the slope by differentiating the function and substituting the given value of x to find the rate of elevation change.
Derive the second derivative of 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)]
Using the chain rule: d²y/dx² = -2x/(1+x²)²
Thus, the second derivative is -2x/(1+x²)².
We calculate the second derivative by differentiating the first derivative using the chain rule and simplifying the expression.
Prove: d/dx (tan⁻¹(x²)) = 2x/(1+x⁴).
Apply the chain rule: Let y = tan⁻¹(x²).
Then dy/dx = (d/dx [x²])/(1+(x²)²) = (2x)/(1+x⁴).
Thus, d/dx (tan⁻¹(x²)) = 2x/(1+x⁴), proving the statement.
The chain rule is used to differentiate the composite function, and the derivative of the inner function is substituted to derive the final result.
Solve: d/dx (tan⁻¹(3x)).
Differentiate using the chain rule: d/dx (tan⁻¹(3x)) = (3)/(1+(3x)²) = 3/(1+9x²). Thus, d/dx (tan⁻¹(3x)) = 3/(1+9x²).
We apply the chain rule to differentiate the function, considering the derivative of the inner function 3x and simplifying to obtain the result.
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