/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 \(7-42\) " Find the derivative o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(7-42\) " Find the derivative of the function. $$f(x)=\frac{1}{(1+\sec x)^{2}}$$

Short Answer

Expert verified
The derivative is \( f'(x) = -2 \sec x \tan x (1 + \sec x)^{-3} \).

Step by step solution

01

Rewrite the Function

We start by rewriting the function in a more convenient form for differentiation. The given function is \( f(x) = \frac{1}{(1 + \sec x)^2} \). We can rewrite this as \( f(x) = (1 + \sec x)^{-2} \).
02

Differentiate Using the Chain Rule

To find the derivative \( f'(x) \), we'll apply the chain rule. The derivative of \( u(x)^n \) where \( u(x) = 1 + \sec x \) and \( n = -2 \) is \( n \cdot u(x)^{n-1} \cdot u'(x) \). Thus, \( \frac{d}{dx}( (1 + \sec x)^{-2} ) = -2(1 + \sec x)^{-3} \cdot (\frac{d}{dx}(1 + \sec x)) \).
03

Differentiate \(1 + \sec x\)

Now, differentiate \(1 + \sec x\) with respect to \(x\). The derivative of \(1\) is \(0\), and the derivative of \(\sec x\) is \(\sec x \tan x\). So, \( \frac{d}{dx}(1 + \sec x) = \sec x \tan x \).
04

Apply the Chain Rule Result

Substitute \( \sec x \tan x \) back into the equation from Step 2. Hence, \( f'(x) = -2(1 + \sec x)^{-3} \cdot \sec x \tan x \). Simplifying gives us the derivative: \( f'(x) = -2 \sec x \tan x (1 + \sec x)^{-3} \).
05

Final Simplification

Combine the terms for the final expression of the derivative, if needed. The derivative is \( f'(x) = -2 \sec x \tan x \cdot (1 + \sec x)^{-3} \), and that's as simple as it gets given the components.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chain Rule
The chain rule is an essential principle in calculus used to differentiate composite functions.It states that if you have a function composed of two functions, say \( f(g(x)) \), then the derivative is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.Simply put, it helps you "chain" together derivatives of layered functions.

In practice, for a function \( u^n \) where \( u = g(x) \) and \( n \) is a constant, the chain rule tells us:
  • Take the derivative of the outer function \( u^n \) to get \( n \cdot u^{n-1} \).
  • Then multiply by the derivative of the inner function \( u' \).
Applying the chain rule allows you to handle complex expressions, like those including powers or trigonometric functions.For our exercise, it was vital in differentiating the function \( f(x) = (1 + \sec x)^{-2} \).
Trigonometric Derivatives
Trigonometric functions have specific derivatives that often come up in calculus.Understanding these can make handling more intricate functions simpler.

For instance, the derivative of \( \sec x \) is \( \sec x \tan x \).This happens because \( \sec x = \frac{1}{\cos x} \), and we're applying derivatives rules like the quotient rule subtly here.Knowing the derivatives of basic trigonometric functions streamlines the process of differentiation.When we encounter something like \( 1 + \sec x \), breaking it into simpler parts helps, since the derivative of 1 is just zero.

In the given exercise, understanding these trigonometric derivatives helped simplify the differentiation of the nested function.
Function Simplification
Simplifying functions makes calculus problems more manageable and often prevents errors.By rewriting functions in a different form, like transforming \( \frac{1}{(1 + \sec x)^2} \) into \( (1 + \sec x)^{-2} \), you can represent the problem in an easier-to-differentiate form.

This technique is especially useful when dealing with powers since power rules are straightforward to apply, compared with more complex fraction derivatives.It also reduces mistakes during differentiation steps, like forgetting to square a denominator or missing out on an entire term.
Remember, the goal is clarity.Reworking complicated parts into simpler terms opens the door to analytical insights, enabling smooth application of rules like the chain rule, as shown in our exercise.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A spherical balloon is being inflated. Find the rate of increase of the surface area \(\left(S-4 \pi r^{2}\right)\) with respect to the radius \(r\) when \(r\) is (a) 1 \(\mathrm{ft}\) , (b) 2 \(\mathrm{ft}\) , and (c) 3 ft. What conclusion can you make?

The cost function for production of a commodity is $$C(x)=339+25 x-0.09 x^{2}+0.0004 x^{3}$$ (a) Find and interpret \(C^{\prime}(100) .\) (b) Compare \(C^{\prime}(100)\) with the cost of producing the 101 st item.

Use the Chain Rule to show that if \(\theta\) is measured in degrees, then $$ \frac{d}{d \theta}(\sin \theta)=\frac{\pi}{180} \cos \theta $$ (This gives one reason for the convention that radian measure is always used when dealing with trigonometric functions in calculus: The differentiation formulas would not be as simple if we used degree measure.)

One side of a right triangle is known to be 20 \(\mathrm{cm}\) long and the opposite angle is measured as \(30^{\circ},\) with a possible error of \(\pm 1^{\circ} .\) (a) Use differentials to estimate the error in computing the length of the hypotenuse. (b) What is the percentage error?

Newton's Law of Gravitation says that the magnitude \(F\) of the force exerted by a body of mass \(m\) on a body of mass \(M\) is $$F=\frac{G m M}{r^{2}}$$ where \(G\) is the gravitational constant and \(r\) is the distance between the bodies. (a) Find \(d F / d r\) and explain its meaning. What does the minus sign indicate? (b) Suppose it is known that the earth attracts an object with a force that decreases at the rate of 2 \(\mathrm{N} / \mathrm{km}\) when \(r=20,000 \mathrm{km} .\) How fast does this force change when \(r=10,000 \mathrm{km}\) ?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.