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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

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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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
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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