For the function h(x) = −2x³ - 4x + 3√√x, calculate h'(x), h''(x), and h'''(x) . Show all steps. (2) h(x) = -6x ² - 4 + 3 h"(x) = (+6x²-4+3 2√x h(x) = -2x 3-4x + 3√3 dx 2√x- In "(x) = £ (4x²)_ _d (4) + 1/ ² dx dx h"(x) = -60 • 2x+0=3x2 h'(x) = a (-2x ²³-4x + 3√√x) d dx |n₁(x) = d (-2x ³) + d (-2x ³) + d (-4x) + d (-4x) + d dx dx h'(x) = (2) (3x²+4+3 (2) |h'(x) = -6x² - 4 + -_-³² 3 2√√x (3√x) 1/2 3|hx) =12x-3 hly--12x. LIX 4x5* n(x) = - (-3) - (4x√7) - d (4x²√x) • (-3) + - 4• ½ x ¹². -12 dx 4 4X7 4x 认 -6.1X (2 h"(x)=-6-3x • 2 (₁7) (2√x)2 h'(x=-12x-3) 4x√x
For the function h(x) = −2x³ - 4x + 3√√x, calculate h'(x), h''(x), and h'''(x) . Show all steps. (2) h(x) = -6x ² - 4 + 3 h"(x) = (+6x²-4+3 2√x h(x) = -2x 3-4x + 3√3 dx 2√x- In "(x) = £ (4x²)_ _d (4) + 1/ ² dx dx h"(x) = -60 • 2x+0=3x2 h'(x) = a (-2x ²³-4x + 3√√x) d dx |n₁(x) = d (-2x ³) + d (-2x ³) + d (-4x) + d (-4x) + d dx dx h'(x) = (2) (3x²+4+3 (2) |h'(x) = -6x² - 4 + -_-³² 3 2√√x (3√x) 1/2 3|hx) =12x-3 hly--12x. LIX 4x5* n(x) = - (-3) - (4x√7) - d (4x²√x) • (-3) + - 4• ½ x ¹². -12 dx 4 4X7 4x 认 -6.1X (2 h"(x)=-6-3x • 2 (₁7) (2√x)2 h'(x=-12x-3) 4x√x
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter3: Linear And Nonlinear Functions
Section: Chapter Questions
Problem 25MCQ
Related questions
Question
![steps.
For the function h(x) = −2x³ - 4x + 3√x, calculate h'(x), h'' (x), and h''' (x) . Show all
2 h(x) = -6x² - 4+ 3
ī
h"(x) = 4 (+6x²-4 + 3 =
2√x
h(x) = -2x 3-4x +3√3
dx
2√x
h "(x) = 2 ( 6x²)_ _d (4) + d( 2
dx
dx
h"(x) = -60 • 2x+0-3x21
2.
h²(x)= x (-2x ³ - 4x + 3√x
2
3
n'(x) = d (-2x ³) + d (-4x) + d (3√x)
dx
dx
h'(x) = (2) (3x²+4+3 (1)
2
|h²(x) = -6x ² - 4 + ²³
3
25
2√√x
√
3|hx)=12x-3
4x-√7
hly--12x
n(x) = d (~3) • (4x√7) - d (4x√x) = (-3) 4 = 4• zx
: -12
dx
4XXX
4x
-3
x 1
4x X2
4/2
Ex
-6.1X
(2
h" (x) = -67-3x • 2 (₁x)
(2√x)2
h(x=-12x-3)
-3
4x√x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b01f764-cc95-4783-8fec-cdda5efc483b%2F7f377660-ab93-4b03-b8e3-1cdf0577c310%2Fgn4lhk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:steps.
For the function h(x) = −2x³ - 4x + 3√x, calculate h'(x), h'' (x), and h''' (x) . Show all
2 h(x) = -6x² - 4+ 3
ī
h"(x) = 4 (+6x²-4 + 3 =
2√x
h(x) = -2x 3-4x +3√3
dx
2√x
h "(x) = 2 ( 6x²)_ _d (4) + d( 2
dx
dx
h"(x) = -60 • 2x+0-3x21
2.
h²(x)= x (-2x ³ - 4x + 3√x
2
3
n'(x) = d (-2x ³) + d (-4x) + d (3√x)
dx
dx
h'(x) = (2) (3x²+4+3 (1)
2
|h²(x) = -6x ² - 4 + ²³
3
25
2√√x
√
3|hx)=12x-3
4x-√7
hly--12x
n(x) = d (~3) • (4x√7) - d (4x√x) = (-3) 4 = 4• zx
: -12
dx
4XXX
4x
-3
x 1
4x X2
4/2
Ex
-6.1X
(2
h" (x) = -67-3x • 2 (₁x)
(2√x)2
h(x=-12x-3)
-3
4x√x
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