/*! 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 27 Find \(d p / d q\). $$p=5+\fra... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 27

Find \(d p / d q\). $$p=5+\frac{1}{\cot q}$$

Short Answer

Expert verified
\( \frac{d p}{d q} = \sec^2 q \).

Step by step solution

01

Rewrite the Expression in Simplified Form

The expression given is \( p = 5 + \frac{1}{\cot q} \). We know \( \frac{1}{\cot q} = \tan q \), so we can rewrite the expression as \( p = 5 + \tan q \).
02

Differentiate with Respect to q

Now, we need to find the derivative of \( p \) with respect to \( q \). The expression is now \( p = 5 + \tan q \). The derivative of 5 with respect to \( q \) is 0, and the derivative of \( \tan q \) with respect to \( q \) is \( \sec^2 q \). Therefore, \( \frac{d p}{d q} = \sec^2 q \).

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 91影视!

Key Concepts

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

Derivative
A derivative is a fundamental concept in calculus that represents how a function changes as its input changes. Picture a curve on a graph; the derivative at any given point is the slope of the tangent line to the curve at that point.
  • It tells us how steep the curve is or how fast the function value is changing at that point.
  • In mathematical terms, if you have a function \( f(x) \), the derivative, \( f'(x) \) or \( \frac{d}{dx} f(x) \), gives the rate of change of \( f \) concerning \( x \).
Understanding derivatives is crucial because they are used to solve problems involving rates of change in physics, engineering, economics, and many other fields. When you're asked to find \( \frac{dp}{dq} \) in a problem, you are essentially finding out how \( p \) changes as \( q \) changes. This gives useful insights, especially in dynamic systems.
Trigonometric Functions
Trigonometric functions play a crucial role in calculus, especially when studying periodic phenomena such as sound waves, light waves, and tides. They include basic functions like sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)).
  • These functions relate the angles and sides of triangles, and they repeat their values in a regular pattern.
  • Tangent, as used in our exercise example, is computed as \( \tan q = \frac{\sin q}{\cos q} \).
In the given problem, \( p \) included the term \( \frac{1}{\cot q} \), which is the same as \( \tan q \). Remembering these identities can simplify many calculus problems by rewriting expressions in forms that are easier to differentiate. Therefore, understanding these functions and their identities is vital because they help in simplifying and solving many calculus problems efficiently.
Differentiation Rules
Differentiation rules are formulas that allow us to find derivatives quickly and accurately. Knowing these rules by heart can help solve complex problems much faster.
  • The power rule allows you to differentiate a term like \( x^n \) easily by bringing the power down and subtracting one from the exponent.
  • The sum rule lets you differentiate terms separately when they are added or subtracted inside a function.
  • For trigonometric functions, specific differentiation rules apply: the derivative of \( \tan q \) is \( \sec^2 q \).
In the step-by-step solution, we used the differentiation rule for tangent. The rule states that the derivative of \( \tan q \) is \( \sec^2 q \), which is handy to remember for exercises like ours. Mastering these rules helps make the process of finding derivatives straightforward and efficient.

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

You will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\). c. Find an equation for the tangent line to \(f\) at the specified $$ \text { point }\left(x_{0}, f\left(x_{0}\right)\right) $$ d. Find an equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 3 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g\), the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal (the line \(y=x\) ). $$y=\frac{x^{3}}{x^{2}+1}, \quad-1 \leq x \leq 1, \quad x_{0}=1 / 2$$

Find the derivative of \(y\) with respect to the given independent variable. $$y=3^{\log _{2} t}$$

Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\sqrt[3]{\frac{x(x-2)}{x^{2}+1}}$$

The linearization of \(2^{x}\) a. Find the linearization of \(f(x)=2^{x}\) at \(x=0 .\) Then round its coefficients to two decimal places. b. Graph the linearization and function together for \(-3 \leq x \leq 3\) and \(-1 \leq x \leq 1\)

Write a differential formula that estimates the given change in volume or surface area. The change in the surface area \(S=6 x^{2}\) of a cube when the edge lengths change from \(x_{0}\) to \(x_{0}+d x\)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation
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.