Problem

(a) If $ f(x) = (x^2 - 1) e^x, $ find $ f' (x) $ …

02:21

Problem 39 Hard Difficulty

(a) If $ f(x) = e^x/ (2x^2 + x + 1), $ find $ f' (x). $
(b) Check to see that your answers to part (a) are reasonable by comparing the graphs of $ f, f', $ and $ f". $

Answer

(a) $f^{\prime}(x)=\frac{4 x}{\left(x^{2}+1\right)^{2}}$
$f^{\prime \prime}(x)=\frac{4\left(1-3 x^{2}\right)}{\left(x^{2}+1\right)^{3}}$
(b) See solution

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Video Transcript

he It's clears the when you read here. So you have up of X is equal to X square minus one over X square plus one. We're gonna use a quotient role. To find our derivative, we get X square plus one D over deep rex of X square minus one minus X square minus one D over DX of X square plus one all over X square plus one square has becomes equal to X square plus one arms to X minus X square minus one times two x over X square plus one square. This becomes equal to four X over X squared, plus one square. Now we're gonna apply the same rule to find the second derivative when we get X Square plus one square D over D X for X minus for X times D over DX for X square plus one square all over X square plus one square and you score that again and then this simplifies into four minus 12 X square over X square plus one cubed. Here you apply the chain rule. Now we're going to draw some graphs for part B. We're going to start with R F of X so we'll draw down in black, an uncle like this, and our red is gonna be our derivative first derivative, and then our second derivative is gonna be in green.

Clarissa N.

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Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 2

The Product and Quotient Rules

Problem

(a) If $ f(x) = (x^2 - 1) e^x, $ find $ f' (x) $ …

Problem 39 Hard Difficulty

(a) If $ f(x) = e^x/ (2x^2 + x + 1), $ find $ f' (x). $
(b) Check to see that your answers to part (a) are reasonable by comparing the graphs of $ f, f', $ and $ f". $

Answer

(a) $f^{\prime}(x)=\frac{4 x}{\left(x^{2}+1\right)^{2}}$
$f^{\prime \prime}(x)=\frac{4\left(1-3 x^{2}\right)}{\left(x^{2}+1\right)^{3}}$
(b) See solution

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Calculus: Early Transcendentals

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Video Transcript

Calculus: Early Transcendentals

Grace H.

Catherine R.

Caleb E.

Kristen K.

Precalculus Review - Intro

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(a) $f^{\prime}(x)=\frac{4 x}{\left(x^{2}+1\right)^{2}}$ $f^{\prime \prime}(x)=\frac{4\left(1-3 x^{2}\right)}{\left(x^{2}+1\right)^{3}}$(b) See solution

Calculus: Early Transcendentals

Grace H.

Catherine R.

Caleb E.

Kristen K.

Precalculus Review - Intro

Functions on the Real Line - Overview

Grace H.

Catherine R.

Caleb E.

Kristen K.

(a) $f^{\prime}(x)=\frac{4 x}{\left(x^{2}+1\right)^{2}}$
$f^{\prime \prime}(x)=\frac{4\left(1-3 x^{2}\right)}{\left(x^{2}+1\right)^{3}}$
(b) See solution