/*! 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 1 In Exercises \(1-12,\) find \(d ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Exercises \(1-12,\) find \(d y / d x\) $$ y=-10 x+3 \cos x $$

Short Answer

Expert verified
\(\frac{dy}{dx} = -10 - 3\sin x\)

Step by step solution

01

Differentiate the Linear Term

We start by differentiating the first term of the function, \(-10x\). The derivative of a linear term \(-10x\) with respect to \(x\) is simply the coefficient of \(x\). Thus, the derivative is \(-10\).
02

Differentiate the Trigonometric Term

Now, we differentiate the second term, \(3\cos x\). The derivative of \(\cos x\) is \(-\sin x\). Thus, the derivative of \(3\cos x\) is \(3(-\sin x) = -3\sin x\).
03

Combine the Derivatives

Finally, we combine the derivatives of each term that we found in the previous steps. Therefore, the overall derivative is the sum of the derivatives of the terms: \(\frac{dy}{dx} = -10 - 3\sin x\).

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.

The Basics of Derivatives
Understanding derivatives is crucial in calculus. They represent the rate at which a function changes as its input changes. If you imagine driving a car, the derivative of your distance over time is your speed, showing how fast you're going. In mathematical terms, the derivative of a function \(f(x)\) with respect to \(x\) is denoted by \(f'(x)\) or \(\frac{df}{dx}\). This notation reflects the "change in \(f\) with respect to change in \(x\)."For linear functions like \(-10x\), the derivative is simply the coefficient of \(x\). This is because the slope, which the derivative represents, is constant. For more complex or non-linear functions, however, finding the derivative involves more intricate rules and methods.
Introduction to Trigonometric Derivatives
Trigonometric derivatives are an essential part of calculus, particularly when dealing with functions that include trigonometric components such as \(\sin x\), \(\cos x\), and \(\tan x\). Understanding these derivatives helps us analyze how trigonometric functions change over different inputs. Here are a few key derivatives:
  • The derivative of \(\sin x\) is \(\cos x\).
  • The derivative of \(\cos x\) is \(-\sin x\).
  • The derivative of \(\tan x\) is \(\sec^2 x\).
In our original problem, we used the derivative of \(\cos x\), which is \(-\sin x\). So, when differentiating \(3\cos x\), we multiply the constant \(3\) by \(-\sin x\), resulting in the term \(-3\sin x\). This process demonstrates how the derivative behaves similarly to a scaling factor for the function.
A Dive into Calculus and its Applications
Calculus is fundamentally about change and accumulation. Derivatives are the tool that calculus provides for examining the first part of this process—understanding change. They allow us to determine how a function behaves at any point, offering insights into trends, maxima, minima, and more. In practical terms:
  • Physics uses calculus to describe motion through the derivatives as velocity and acceleration.
  • Economics uses them to find optimal production levels.
  • Biology uses them for modeling population changes.
In solving calculus problems, it's important to apply differentiation rules methodically, as seen with our example. Recognizing constant multiples and trigonometric identities, like we did, is part of becoming proficient in calculus. It makes solving complex equations more manageable and provides essential insights, whether for scientific research or everyday calculations.

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

Show that the approximation of \(\sqrt{1+x}\) by its linearization at the origin must improve as \(x \rightarrow 0\) by showing that $$ \lim _{x \rightarrow 0} \frac{\sqrt{1+x}}{1+(x / 2)}=1 $$

In Exercises \(87-94\) , find an equation for the line tangent to the curve at the point defined by the given value of \(t .\) Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=t-\sin t, \quad y=1-\cos t, \quad t=\pi / 3 $$

Tangents Suppose that \(u=g(x)\) is differentiable at \(x=1\) and that \(y=f(u)\) is differentiable at \(u=g(1)\) . If the graph of \(y=f(g(x))\) has a horizontal tangent at \(x=1,\) can we conclude anything about the tangent to the graph of \(g\) at \(x=1\) or the tangent to the graph of \(f\) at \(u=g(1) ?\) Give reasons for your answer.

The linearization is the best linear approximation This is why we use the linearization.) Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the erroo \(E(x)=f(x)-g(x)\) were small enough near \(x=a,\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=\) \(f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions 1\. \(E(a)=0\) 2\. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\)

In Exercises \(47-56,\) verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$ 6 x^{2}+3 x y+2 y^{2}+17 y-6=0, \quad(-1,0) $$

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.