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Chapter 5: Problem 5

Find the derivative of \(y=f(x)=\tan ^{-1} x\), using first principle of differentiation.

Short Answer

Expert verified
The derivative of the function \(y=f(x)=\tan ^{-1} x\) or arctanx using the first principles of differentiation is 1

Step by step solution

01

Write out formula for differentiation from first principles

The formula for differentiation using first principles is \(f'(x)=\lim_{{h\to 0}}\frac{f(x+h)-f(x)}{h}\). This will be used to find the derivative of the function \(y=f(x)=\tan ^{-1} x\).
02

Apply the formula to our function

Taking \(f(x)=\tan ^{-1} x\), we apply the formula we identified in step 1. So, \(f'(x)=\lim_{{h\to 0}}\frac{\tan^{-1}(x+h) - \tan^{-1}(x)}{h}\). Now we have all the components in place for solving the problem.
03

Use the addition formula for \(\tan ^{-1}\)

We know that \(\tan^{-1}(a+b) = \tan^{-1}(a) + \tan^{-1}(b)\) only when \(ab < 1\), but we cannot guarantee this for our case because we do not know the values of a and b. Therefore, we use the formula \(\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right)\), and rewrite our expression: \(f'(x)=\lim_{{h\to 0}}\frac{\tan^{-1}(x)+\tan^{-1}(h) - \tan^{-1}(x)}{h} =\lim_{{h\to 0}}\frac{\tan^{-1}(h)}{h}\).
04

Simplify

We know that for very small values of h, \(\tan^{-1}(h)\) is approximately equal to h. Hence, we get \(f'(x)=\lim_{{h\to 0}}\frac{h}{h} = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Trigonometric Functions
Inverse trigonometric functions are vital in calculus because they allow us to find angles when given trigonometric values. The function \( \tan^{-1}(x) \), also known as arctan, gives the angle whose tangent is \( x \). These functions are essential in fields such as geometry and engineering, especially when determining angles in right triangles.
Unlike regular trigonometric functions, which can have infinite values, inverse functions are restricted to specific ranges:
  • \( \tan^{-1}(x) \) outputs angles between \( -\frac{\pi}{2} \) and \( \frac{\pi}{2} \).
  • The behavior of inverse functions near boundaries is crucial for differentiation and integration.
Understanding how these functions operate helps when differentiating them, as different rules apply compared to standard trigonometric functions.
Limits
Limits describe how a function behaves as it approaches a particular input. They're the backbone of solving many calculus problems, especially when determining derivatives via the first principle.
In our exercise, we use limits to find the derivative of \( f(x) = \tan^{-1}(x) \). The approach is:
  • Set up the limit \( \lim_{{h\to 0}}\frac{f(x+h)-f(x)}{h} \) to find the instantaneous rate of change.
  • Apply known identities like \( \tan^{-1}(a+b) = \tan^{-1}(a) + \tan^{-1}(b) \) for simplification, although adjusted here due to constraints.
  • Simplify using approximations. For small \( h \), \( \tan^{-1}(h) \) is approximately \( h \), leading to a simpler result.
Understanding limits helps in breaking down complex expressions and finding derivatives accurately.
Calculus
Calculus is a broad field that focuses on changes and motion. Two primary operations in calculus are differentiation and integration. Differentiation helps us determine how a function shifts as its input alters, which is achieved using derivatives.
Using the first principle of differentiation:
  • The formula \( f'(x)=\lim_{{h\to 0}}\frac{f(x+h)-f(x)}{h} \) lets us find derivatives directly by understanding how the function behaves as \( h \) becomes negligible.
  • This principle requires a deep dive into function properties, making it foundational for more advanced calculus concepts.
In practical terms, calculus lets us model real-world phenomena, solve optimization problems, and analyze variable rates鈥攕erving as the foundation for countless scientific and engineering applications.

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Most popular questions from this chapter

If \(y=\cot ^{-1}\left\\{\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right\\}\), then prove that \(\frac{d y}{d x}\) is independent of \(x\).

Suppose \(f\) is a differentiable function such that \(f(g(x))=x\) and \(f^{\prime}(x)=1+(f(x))^{2}\), then prove that \(g^{\prime}(x)=\frac{1}{1+x^{2}} .\)

If \(x=a(\theta+\sin \theta), y=a(1+\cos \theta)\) prove that \(\frac{d^{2} y}{d x^{2}}=-\frac{a}{y^{2}}\)

Find \(\frac{d^{2} y}{d x^{2}}\), if (i) \(x=a t^{2}, y=2 a t\) (ii) \(x=a \cos ^{3} \theta, y=a \sin ^{3} \theta\) (iii) \(x=a \cos \theta, y=b \sin \theta\)

If \(y=\left(1+\frac{1}{x}\right)^{x}+x^{\left(1+\frac{1}{x}\right)}\), find \(\frac{d y}{d x}\) at \(x=1\)
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