Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find \(d y / d x\) by implicit differentiation and evaluate the derivative at the indicated point. $$ x \cos y=1, \quad\left(2, \frac{\pi}{3}\right) $$

Short Answer

Expert verified
The derivative of the function at the point \((2, \frac{\pi}{3})\) is \(\sqrt{3}\).

Step by step solution

01

Implicit differentiation

Start by taking the derivative of each side of the equation \(x \cos y = 1\). This results in the equation \((1) \cos y - x \sin y (y') = 0\), where \(y'\) is the derivative of \(y\) with respect to \(x\).
02

Solving for \(y'\)

Rearrange the equation to solve for \(y'\). The rearranged equation is \(y' = \frac{\cos y}{\sin y}\).
03

Evaluation at point

Substitute the given point \((2, \frac{\pi}{3})\) into the equation for \(y'\). This results in \(y'(2, \frac{\pi}{3}) = \frac{\cos \frac{\pi}{3}}{\sin \frac{\pi}{3}}\).
04

Simplify

Simplify the right hand side to obtain the value of the derivative at the point. The simplified value is \(y'(2, \frac{\pi}{3}) = \sqrt{3}\).

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 Vaia!

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

The area of a square with sides of length \(s\) is given by \(A=s^{2} .\) Find the rate of change of the area with respect to \(s\) when \(s=4\) meters.

The normal daily maximum temperatures \(T\) (in degrees Fahrenheit) for Denver, Colorado, are shown in the table. (Source: National Oceanic and Atmospheric Administration). $$ \begin{aligned} &\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Month } & \text { Jan } & \text { Feb } & \text { Mar } & \text { Apr } & \text { May } & \text { Jun } \\ \hline \text { Temperature } & 43.2 & 47.2 & 53.7 & 60.9 & 70.5 & 82.1 \\ \hline \end{array}\\\ &\begin{array}{|l|c|c|c|c|c|c|} \hline \text { Month } & \text { Jul } & \text { Aug } & \text { Sep } & \text { Oct } & \text { Nov } & \text { Dec } \\ \hline \text { Temperature } & 88.0 & 86.0 & 77.4 & 66.0 & 51.5 & 44.1 \\ \hline \end{array} \end{aligned} $$(a) Use a graphing utility to plot the data and find a model for the data of the form \(T(t)=a+b \sin (\pi t / 6-c)\) where \(T\) is the temperature and \(t\) is the time in months, with \(t=1\) corresponding to January. (b) Use a graphing utility to graph the model. How well does the model fit the data? (c) Find \(T^{\prime}\) and use a graphing utility to graph the derivative. (d) Based on the graph of the derivative, during what times does the temperature change most rapidly? Most slowly? Do your answers agree with your observations of the temperature changes? Explain.

Find the average rate of change of the function over the given interval. Compare this average rate of change with the instantaneous rates of change at the endpoints of the interval. $$ g(x)=x^{2}+e^{x}, \quad[0,1] $$

Prove that \(\arcsin x=\arctan \left(\frac{x}{\sqrt{1-x^{2}}}\right),|x|<1\)

A 15 -centimeter pendulum moves according to the equation \(\theta=0.2 \cos 8 t,\) where \(\theta\) is the angular displacement from the vertical in radians and \(t\) is the time in seconds. Determine the maximum angular displacement and the rate of change of \(\theta\) when \(t=3\) seconds.

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.

Sign-up for free