Differentiating Arctan(3x) A Step-by-Step Guide

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Introduction

In this comprehensive guide, we will delve into the process of differentiating the function $f(x) = \tan^{-1}(3x)$. This problem falls under the domain of calculus, specifically dealing with the differentiation of inverse trigonometric functions. Mastering such differentiation techniques is crucial for various applications in physics, engineering, and other scientific fields. Our primary focus will be on applying the chain rule in conjunction with the derivative of the arctangent function to arrive at the exact derivative, $f'(x)$. We will break down each step in a detailed manner, ensuring clarity and understanding for readers with varying levels of calculus proficiency. This article aims to provide not just the solution, but a thorough explanation of the underlying principles, thereby enhancing the reader's problem-solving skills in calculus.

Understanding the Arctangent Function and Its Derivative

The arctangent function, denoted as $ \tan^{-1}(x) $ or $ \arctan(x) $, is the inverse of the tangent function. It returns the angle whose tangent is $ x $. A fundamental concept required for differentiating $ f(x) = \tan^{-1}(3x) $ is the derivative of the arctangent function itself. The derivative of $ \tan^{-1}(x) $ with respect to $ x $ is given by:

ddx[tan⁡−1(x)]=11+x2\frac{d}{dx} [\tan^{-1}(x)] = \frac{1}{1 + x^2}

This formula is a cornerstone in calculus and is derived using implicit differentiation. To fully grasp this, consider $ y = \tan^{-1}(x) $. This implies that $ x = \tan(y) $. Differentiating both sides with respect to $ y $, we get:

dxdy=sec⁡2(y)\frac{dx}{dy} = \sec^2(y)

Since $ \sec^2(y) = 1 + \tan^2(y) $, we can rewrite this as:

dxdy=1+tan⁡2(y)=1+x2\frac{dx}{dy} = 1 + \tan^2(y) = 1 + x^2

Now, to find $ \frac{dy}{dx} $, which is the derivative of $ \tan^{-1}(x) $, we take the reciprocal of $ \frac{dx}{dy} $, yielding:

dydx=11+x2\frac{dy}{dx} = \frac{1}{1 + x^2}

This foundational understanding is crucial because differentiating $ f(x) = \tan^{-1}(3x) $ requires us to apply the chain rule, which builds upon this basic derivative.

Applying the Chain Rule

The chain rule is a pivotal concept in calculus used to differentiate composite functions. A composite function is a function that is composed of another function. Mathematically, if we have a function $ f(g(x)) $, the chain rule states that the derivative of this composite function is given by:

ddx[f(g(x))]=f′(g(x))imesg′(x)\frac{d}{dx} [f(g(x))] = f'(g(x)) imes g'(x)

In simpler terms, the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This rule is essential when dealing with functions within functions, such as our given function, $ f(x) = \tan^{-1}(3x) $. Here, the outer function is $ \tan^{-1}(u) $ and the inner function is $ g(x) = 3x $. Applying the chain rule allows us to systematically differentiate such complex functions by breaking them down into simpler, manageable parts. Understanding and correctly applying the chain rule is paramount for success in calculus, as it appears in various forms and contexts throughout the subject.

Step-by-Step Differentiation of $f(x) = \tan^{-1}(3x)$

To differentiate $ f(x) = \tan^{-1}(3x) $, we will apply the chain rule meticulously. Let's break this down step by step:

  1. Identify the Outer and Inner Functions: In this case, the outer function is $ f(u) = \tan^{-1}(u) $ and the inner function is $ g(x) = 3x $. So, we can express $ f(x) $ as $ f(g(x)) = \tan^{-1}(g(x)) $.

  2. Differentiate the Outer Function: The derivative of $ \tan^{-1}(u) $ with respect to $ u $ is $ \frac{1}{1 + u^2} $. Therefore, $ f'(u) = \frac{1}{1 + u^2} $.

  3. Differentiate the Inner Function: The derivative of $ g(x) = 3x $ with respect to $ x $ is simply 3. So, $ g'(x) = 3 $.

  4. Apply the Chain Rule: Using the chain rule formula, $ \frac{d}{dx} [f(g(x))] = f'(g(x)) imes g'(x) $, we substitute the derivatives we found in the previous steps.

    Thus, we have:

    f′(x)=ddx[tan⁡−1(3x)]=11+(3x)2imes3f'(x) = \frac{d}{dx} [\tan^{-1}(3x)] = \frac{1}{1 + (3x)^2} imes 3

  5. Simplify the Expression: Now, we simplify the expression to obtain the final derivative:

    f′(x)=31+9x2f'(x) = \frac{3}{1 + 9x^2}

Therefore, the derivative of $ f(x) = \tan^{-1}(3x) $ is $ f'(x) = \frac{3}{1 + 9x^2} $. This step-by-step approach not only gives us the correct answer but also reinforces the application of the chain rule in differentiating composite functions. Each step is crucial, from identifying the outer and inner functions to the final simplification, ensuring a thorough understanding of the process.

Final Result: The Derivative of $f(x) = \tan^{-1}(3x)$

After carefully applying the chain rule and simplifying the expression, we arrive at the final derivative of the function $ f(x) = \tan^{-1}(3x) $. The derivative, denoted as $ f'(x) $, is given by:

f′(x)=31+9x2\boxed{f'(x) = \frac{3}{1 + 9x^2}}

This result is obtained by correctly identifying the outer and inner functions, differentiating each separately, and then combining them using the chain rule. The process involved differentiating the arctangent function and then multiplying by the derivative of the inner function, which was $ 3x $. The simplification process then led us to the concise and accurate final form. Understanding this process is vital for mastering differentiation techniques in calculus. This derivative, $ f'(x) = \frac{3}{1 + 9x^2} $, represents the instantaneous rate of change of the function $ f(x) = \tan^{-1}(3x) $ with respect to $ x $.

Conclusion

In conclusion, we have successfully differentiated the function $ f(x) = \tan^{-1}(3x) $ by applying the chain rule. This involved recognizing the function as a composite function, differentiating the outer and inner functions separately, and then combining these derivatives according to the chain rule formula. The final result, $ f'(x) = \frac{3}{1 + 9x^2} $, provides the exact derivative of the given function. The step-by-step methodology employed here is crucial for accurately differentiating complex functions in calculus. This process not only answers the specific question but also reinforces the understanding and application of fundamental calculus principles, such as the chain rule and the derivative of inverse trigonometric functions. Mastering these techniques is essential for further studies in calculus and its applications in various fields such as physics, engineering, and economics. The ability to differentiate such functions accurately is a valuable skill for anyone working in quantitative disciplines.