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If \(f(x)=\tan x,\) show that \(\frac{f(x+h)-f(x)}{h}=\sec ^{2} x\left(\frac{\sin h}{h}\right) \frac{1}{\cos h-\sin h \tan x}\).

Short Answer

Expert verified
This is the expression for the difference quotient of \(f(x) = \tan x\).

Step by step solution

01

Understanding the Problem

We need to find the difference quotient for the function \(f(x) = \tan x\), given by \(\frac{f(x+h) - f(x)}{h}\), and show it equals a specific expression involving trigonometric functions and limits.
02

Express Functional Difference

Start by expressing \(f(x+h) - f(x) = \tan(x + h) - \tan x\). We use the tangent addition formula: \(\tan(x + h) = \frac{\tan x + \tan h}{1 - \tan x \tan h}\). Substitute this into our expression.
03

Substitute Tangent Addition Formula

Replace \(\tan(x + h)\) in the expression: \(f(x+h) - f(x) = \frac{\tan x + \tan h}{1 - \tan x \tan h} - \tan x\). Simplify to get a single fraction: \(= \frac{\tan x + \tan h - \tan x(1 - \tan x \tan h)}{1 - \tan x \tan h}\).
04

Simplify Expression

Simplify further, noticing \(\tan x + \tan h - \tan x + \tan^2 x \tan h = \tan h (1 + \tan^2 x)\). Thus, \(= \frac{\tan h (1 + \tan^2 x)}{1 - \tan x \tan h}\).
05

Express in Terms of Sine and Cosine

Recall \(1 + \tan^2 x = \sec^2 x\). Therefore, \(\frac{\tan h \sec^2 x}{1 - \tan x \tan h}\). Now, express \(\tan h = \frac{\sin h}{\cos h}\). Substitute these into the formula.
06

Simplify Further with Identical Terms

Substitute \(\tan h\): \(\frac{\frac{\sin h}{\cos h} \sec^2 x}{1 - \frac{\sin h \tan x}{\cos h}}\). Factor out \(\sec^2 x\): \(= \sec^2 x \cdot \frac{\sin h}{\cos h - \sin h \tan x}\).
07

Put it in Derivative Form

Divide the expression by \(h\): \(\frac{\sec^2 x \cdot \frac{\sin h}{\cos h - \sin h \tan x}}{h} = \sec^2 x \cdot \left(\frac{\sin h}{h}\right) \cdot \frac{1}{\cos h - \sin h \tan x}\).
08

Final Verification

The expression matches the provided one. Remember, \(\lim_{h \to 0} \frac{\sin h}{h} = 1\), validating the simplification. Therefore, the expression is correct.

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Key Concepts

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

Difference Quotient
The difference quotient is a key concept in calculus that helps us understand the rate of change of a function. When we take the function \(f(x)\) and alter it slightly to \(f(x+h)\), the difference in the function values over this small change in \(x\) gives us valuable information about the behavior of the function.

In this exercise, we're focusing on the function \(f(x) = \tan x\). To find the difference quotient, we use the formula:
  • \(\frac{f(x+h) - f(x)}{h}\)
Our goal involves transforming this expression using trigonometric identities to eventually express it in a way involving \(\sec^2 x\). It is a fundamental step toward understanding derivatives in calculus, showing not only how functions behave locally but also how complex functions can be simplified.
Trigonometric Identities
Trigonometric identities are expressions that hold true for all values of the involved variables.

In this exercise, we extensively use the tangent addition formula:
  • \(\tan(x + h) = \frac{\tan x + \tan h}{1 - \tan x \tan h}\)
  • This formula is key to simplifying \(f(x+h) - f(x)\) when \(f(x) = \tan x\).
Trigonometric identities like \(1 + \tan^2 x = \sec^2 x\) also play a crucial role.
  • Such identities transform expressions, aiding in simplification.
These transformations reduce complex trigonometric expressions into more manageable forms, which is essential in various calculus problems, enabling us to derive meaningful results more efficiently.
Limits in Calculus
Limits are cornerstones of calculus that permit us to make sense of behaviors as an input tends to a certain value.

In problems like the one at hand, the limit concept is employed to match intermediate expressions to their most simplified forms. In our case:
  • \(\lim_{h \to 0} \frac{\sin h}{h} = 1\)
Using this limit is crucial as \(h\) approaches zero, allowing us to effectively use the squeeze theorem to validate expressions and confirm these limiting behaviors accurately. This helps refine the difference quotient into the perfect form needed to connect with derivatives, showing just how vital limits are in endless calculus applications.
Tangent Addition Formula
The tangent addition formula is fundamental in many trigonometry and calculus problems. It provides the ability to express the tangent of a sum of two angles with a known identity:
  • \(\tan(x + h) = \frac{\tan x + \tan h}{1 - \tan x \tan h}\)
This formula, essential in the simplification steps of this exercise, lets us transform expressions involving \(\tan(x + h)\) and \(\tan x\) into a single fraction.

By substituting specific substitutions like \(\tan h = \frac{\sin h}{\cos h}\), we bridge our trigonometric identities with calculus notions. This showcase how important combining these identities is when tackling advanced problems involving limits and being able to navigate comfortably through is fundamental in finding neat and simplified end results.

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Most popular questions from this chapter

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