/*! 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 59 Finding an Equation of a Tangent... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 59

Finding an Equation of a Tangent Line In Exercises \(55-62,\) find an equation of the tangent line to the graph of the function at the given point. $$ f(x)=e^{-x} \ln x, \quad(1,0) $$

Short Answer

Expert verified
The equation of the tangent line to the graph of the function \(y=2^{-x}\) at the point \((-1,2)\) is \(y - 2 = -2\ln(2)(x -(-1))\) which equates to \(y = -2\ln(2)(x + 1) + 2\)

Step by step solution

01

Find the Derivative

Differentiate the function \(y = 2^{-x}\) with respect to x using the chain rule. The derivative of \(f(x) = a^{-x}\) (where 'a' is a constant) is \(-a^{-x} \ln(a)\). So the derivative \(y' = 2^{-x}\ln(2)\)
02

Substituting the Given Point to Find the Slope of the Tangent Line

Plug in the given point \(-1, 2\) into the derivative to find the slope of the tangent line at that point. Replacing x by -1 in the derivative gives \(y' = 2^{-( -1)}\ln(2) = \(-2\ln(2)\)
03

Generating the Equation of the Tangent Line

Use the point-slope form of the line equation \(y - y1 = m(x - x1)\), substituting -1 for x1, 2 for y1, and \(-2\ln(2)\) for m, we get the equation \(y - 2 = -2\ln(2)(x -(-1)) = -2\ln(2)(x + 1)\)

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 Calculation
Understanding the derivative of a function is crucial when you're trying to find the equation of a tangent line. Derivative calculation involves finding the rate at which a function is changing at any point on its graph. For the function given in our exercise, \(y = 2^{-x}\), we calculate the derivative to find out how the function's output value changes as \(x\) changes. Put simply, if you imagine a curve on a graph, the derivative at any point is the slope of a line that just

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

Proof Prove that $$\tanh ^{-1} x=\frac{1}{2} \ln \left(\frac{1+x}{1-x}\right), \quad-1

Prove each differentiation formula. (a) \(\frac{d}{d x}[\arctan u]=\frac{u^{\prime}}{1+u^{2}}\) (b) \(\frac{d}{d x}[\operatorname{arccot} u]=\frac{-u^{\prime}}{1+u^{2}}\) (c) \(\frac{d}{d x}[\operatorname{arcsec} u]=\frac{u^{\prime}}{|u| \sqrt{u^{2}-1}}\) (d) \(\frac{d}{d x}[\operatorname{arccsc} u]=\frac{-u^{\prime}}{|u| \sqrt{u^{2}-1}}\)

The graphs of \(f(x)=\sin x\) and \(g(x)=\cos x\) are shown below. (a) Explain whether the points \(\left(-\frac{\sqrt{2}}{2},-\frac{\pi}{4}\right), \quad(0,0) \quad\) and \(\quad\left(\frac{\sqrt{3}}{2}, \frac{2 \pi}{3}\right)\) lie on the graph of \(y=\arcsin x\). (b) Explain whether the points \(\left(-\frac{1}{2}, \frac{2 \pi}{3}\right), \quad\left(0, \frac{\pi}{2}\right), \quad\) and \(\quad\left(\frac{1}{2},-\frac{\pi}{3}\right)\) lie on the graph of \(y=\operatorname{arcos} x\).

Deriving an Inequality Given \(e^{x} \geq 1\) for \(x \geq 0,\) it follows that $$ \int_{0}^{x} e^{t} d t \geq \int_{0}^{x} 1 d t $$ Perform this integration to derive the inequality $$ \begin{array}{l}{e^{x} \geq 1+x} \\ {\text { for } x \geq 0}\end{array} $$

Find the derivative of the function. \(f(t)=\arcsin t^{2}\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Applied Mathematics

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.