Derivative Of Arctan Sqrt X

Unveiling the Derivative of arctan(√x): A Comprehensive Guide

Finding the derivative of inverse trigonometric functions can sometimes feel like navigating a mathematical maze. This article provides a thorough exploration of the derivative of arctan(√x), guiding you through the process step-by-step, explaining the underlying principles, and addressing common questions. Understanding this derivative is crucial for various applications in calculus, particularly in solving optimization problems and analyzing curves. We'll cover the chain rule, the derivative of arctan(u), and provide ample explanations to solidify your understanding.

I. Introduction: Understanding the Problem

Our goal is to find the derivative of the function f(x) = arctan(√x). This requires a solid grasp of the chain rule and the derivative of the arctangent function. The chain rule is essential because we are dealing with a composite function: the arctangent function acting on the square root function. Let's break down the problem and then systematically solve it. We will also explore the domain and range of the original function and its derivative to provide a complete understanding.

II. Prerequisites: Key Concepts

Before diving into the derivation, let's refresh some crucial calculus concepts:

• The Chain Rule: If we have a composite function y = f(g(x)), then its derivative is given by dy/dx = f'(g(x)) * g'(x). In simpler terms, we differentiate the "outer" function, leaving the "inner" function untouched, and then multiply by the derivative of the "inner" function.

• The Derivative of arctan(u): The derivative of the arctangent function with respect to u is given by d(arctan(u))/du = 1/(1 + u²). This is a fundamental result that forms the basis of our solution.

• The Power Rule: The derivative of xⁿ is nxⁿ⁻¹. This rule will be used when differentiating the inner function, √x.

III. Step-by-Step Derivation: Finding the Derivative

Now, let's find the derivative of f(x) = arctan(√x) using the chain rule:

1. Identify the inner and outer functions:

   • Outer function: f(u) = arctan(u) where u = √x
   • Inner function: g(x) = √x = x^(1/2)

2. Find the derivative of the outer function:

   • f'(u) = 1/(1 + u²)

3. Find the derivative of the inner function:

   • g'(x) = (1/2)x^(-1/2) = 1/(2√x)

4. Apply the chain rule:

   • f'(x) = f'(g(x)) * g'(x) = [1/(1 + (√x)²)] * [1/(2√x)]

5. Simplify the expression:

   • f'(x) = [1/(1 + x)] * [1/(2√x)] = 1/(2√x(1 + x))

Therefore, the derivative of arctan(√x) is 1/(2√x(1 + x)).

IV. Explanation and Interpretation

The derivative we derived, 1/(2√x(1 + x)), represents the instantaneous rate of change of the function arctan(√x) at any given point x. It tells us how steeply the arctan(√x) curve is rising or falling at that point. Notice that the derivative is only defined for x > 0, reflecting the fact that the square root of a negative number is not a real number and hence the arctan function is not defined for negative x.

The denominator, 2√x(1 + x), plays a crucial role in determining the magnitude and sign of the derivative. As x increases, both √x and (1 + x) increase, causing the denominator to grow larger. This results in a smaller derivative, meaning the rate of change of arctan(√x) decreases as x increases.

The presence of the square root in the denominator indicates that the derivative will be larger for smaller values of x, reflecting a steeper slope at the beginning of the curve. As x approaches infinity, the derivative approaches zero, indicating that the curve flattens out.

V. Domain and Range of the Function and its Derivative

• Original Function f(x) = arctan(√x):

  • Domain: x ≥ 0 (since the square root is only defined for non-negative numbers).
  • Range: [0, π/2) (arctan has a range of (-π/2, π/2), but since √x is always non-negative, the range is restricted to the positive values).

• Derivative f'(x) = 1/(2√x(1 + x)):

  • Domain: x > 0 (the denominator cannot be zero, and the square root is only defined for positive x).
  • Range: (0, ∞) The derivative is always positive indicating that the original function is strictly increasing across its domain.

VI. Illustrative Example

Let's consider a specific point, say x = 1. The value of the function at x = 1 is arctan(√1) = arctan(1) = π/4. The derivative at x = 1 is 1/(2√1(1 + 1)) = 1/4. This means that at x = 1, the tangent line to the curve of arctan(√x) has a slope of 1/4.

VII. Applications

Understanding the derivative of arctan(√x) has practical applications in various fields:

• Physics: Modeling the behavior of certain physical systems that involve inverse trigonometric functions.

• Engineering: Solving optimization problems where the objective function involves arctangent.

• Computer Graphics: Used in algorithms for calculating angles and rotations.

• Statistics: May appear in the context of probability distributions.

VIII. Frequently Asked Questions (FAQ)

• Q: Why is the chain rule necessary here?

  • A: The chain rule is essential because we're dealing with a composite function. The arctangent function is applied to the square root function, making it a nested function that requires the chain rule for differentiation.

• Q: What if I have arctan(√(ax + b))? How would the derivative change?

  • A: You'd still use the chain rule. The inner function becomes √(ax + b), and its derivative would be a/(2√(ax + b)) according to the chain rule. The derivative would then become [a/(2√(ax+b)(1 + (ax+b)))]

• Q: Can I use implicit differentiation to solve this?

  • A: While technically possible, it would be far more complex than using the chain rule directly. The chain rule offers a much more straightforward approach in this particular case.

• Q: What is the significance of the domain restriction (x > 0) for the derivative?

  • A: The domain restriction ensures that the denominator of the derivative doesn't become zero and that the square root is defined for real numbers. Attempting to calculate the derivative at x ≤ 0 would lead to undefined results.

IX. Conclusion

This comprehensive guide has provided a step-by-step derivation of the derivative of arctan(√x), explaining the underlying mathematical principles and offering valuable insights. The derivative, 1/(2√x(1 + x)), reveals the rate of change of the arctan(√x) function and its behavior across its domain. Understanding this derivative opens doors to solving complex problems in calculus and its applications across various fields. Remember to always check the domain and range of both the original function and its derivative for a complete understanding. By mastering this concept, you'll gain a deeper appreciation for the power and elegance of calculus.