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Chapter 3: Problem 8

In Exercises find \(f^{\prime}(x)\). $$f(x)=\left(x^{2}+1\right) \sec x$$

Short Answer

Expert verified
The derivative is \(f'(x) = 2x \sec x + (x^2 + 1) \sec x \tan x\).

Step by step solution

01

Identify the Functions Involved

We have a product of two functions: \( u(x) = x^2 + 1 \) and \( v(x) = \sec x \). We need to differentiate \(f(x) = u(x)v(x)\).
02

Apply the Product Rule

The product rule states that the derivative of a product \(u(x)v(x)\) is given by \(f'(x) = u'(x)v(x) + u(x)v'(x)\). We will differentiate \(u(x)\) and \(v(x)\) separately.
03

Differentiate \(u(x) = x^2 + 1\)

The derivative of \(u(x) = x^2 + 1\) with respect to \(x\) is \(u'(x) = 2x\).
04

Differentiate \(v(x) = \sec x\)

The derivative of \(v(x) = \sec x\) with respect to \(x\) is \(v'(x) = \sec x \tan x\).
05

Substitute Derivatives into Product Rule

Using the product rule formula, substitute in the derivatives: \[f'(x) = (2x)(\sec x) + (x^2 + 1)(\sec x \tan x)\]
06

Simplify the Expression

Simplify the expression from the product rule: \[f'(x) = 2x \sec x + (x^2 + 1) \sec x \tan x\]. Keep it in this form, as it's already expanded.

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Key Concepts

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

Product Rule in Differentiation
Differentiation often deals with a wide range of functions, including those that are products of simpler functions. The product rule is a key differentiation technique that helps in finding the derivative of a product of two functions. When you have two functions, say \( u(x) \) and \( v(x) \), their product \( f(x) = u(x)v(x) \) can be differentiated using the product rule. The rule gives us that the derivative \( f'(x) \) is:
  • \( f'(x) = u'(x)v(x) + u(x)v'(x) \)
This means:
  • First, differentiate the first function \( u(x) \), multiply it by the second function \( v(x) \).
  • Then, differentiate the second function \( v(x) \), multiply it by the first function \( u(x) \).
  • Sum these two results together.
Remembering and applying this rule is essential when dealing with complex expressions where two functions are being multiplied.
Understanding the Derivative of Secant
The function \( \sec x \), the secant function, is one that appears frequently in calculus. Its derivative has a specific form that is useful to remember:
  • \( \frac{d}{dx} [\sec x] = \sec x \tan x \)
This derivative arises from the relationship between secant and other trigonometric functions like cosine and tangent:
  • Recall \( \sec x = \frac{1}{\cos x} \).
  • This means the derivative can initially be found using the quotient rule but simplifies to \( \sec x \tan x \).
Knowing this derivative is key when working with problems involving trigonometric functions, especially in product rules involving terms such as \( \sec x \). Keeping this formula in mind speeds up solving derivatives of such trigonometric products.
Differentiation Techniques Overview
The process of differentiation is foundational in calculus and involves finding how a function changes as its input changes. Differentiation techniques help solve a wide range of problems:
  • The product rule is essential for functions that are products of other functions, as seen in the provided exercise.
  • The chain rule is used when differentiating compositions of functions.
  • The quotient rule applies when you have a function that is a division of two others.
Alongside these, trigonometric derivatives, like the derivative of \( \sec x \), often combine with these rules. Each technique arises from the basic principle of limits and function behavior. Grasping these fundamental techniques allows one to tackle almost any differentiation problem, leading to a deeper understanding of changes and rates of change in various problems.

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Most popular questions from this chapter

In each part, determine where \(f\) is differentiable. (a) \(f(x)=\sin x\) (b) \(f(x)=\cos x\) (c) \(f(x)=\tan x\) (d) \(f(x)=\cot x\) (e) \(f(x)=\sec x\) (f) \(f(x)=\csc x\) \((g) f(x)=\frac{1}{1+\cos x}\) (h) \(f(x)=\frac{1}{\sin x \cos x}\) (i) \(f(x)=\frac{\cos x}{2-\sin x}\)

Find formulas for \(d y\) and \(\Delta y\) at a general point \(x\). $$y=\sin x$$

Find \(f^{\prime}(x).\) $$f(x)=\left(3 x^{2}+1\right)^{2}$$

find \(d y / d x\) $$y=\sin (\tan 3 x)$$

find \(f^{\prime}(x)\) $$f(x)=\sqrt{4+\sqrt{3 x}}$$
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