Use your knowledge on Higher-order derivative and Chain Rule to show that z" + 4z' + 8z = 0 if z = e-2x (sin 2x + cos 2x). Z = e-2x (sin 2x + cos 2x) Let u = 2x, then du = 2dx du = 2 dx Z = e-"(sin u + cos u) z'%3Dе"Dx(sin и + cos u) + (sin и + cos u)Dx(е ") z'%3Dе " (сos u — sin u) — е (sinu + cosu) coS и — е "и sin u — e"u sin u — e"и cos u dy = -2e¬u sin u du dy = -2e¬u sin u, du du Substitute = 2 using the Chain Rule, dx dz dy du (-2e¬" sin u)2 = -4e¬u sin u du du dx Substitute u = 2x, z' = -4e-2x sin 2x To get the second derivative, again let u = 2x then du = 2dx du = 2 dx z' = -4e-u sin u -4[e-"Dx(sin u) + sin u Dx(e¬u) 3 -4е И cos и — 4е -И sin u z" = dy — — 4е и соs и — 4e-U sin u du dy = -4e-u du Substitute cOS и — 4e"u sin u, = 2 using the Chain Rule, dx du dz dy du (-4е " cos u — 4е И sin u)2 3D —8е И сosu — 8e u sin u du du dx
Use your knowledge on Higher-order derivative and Chain Rule to show that z" + 4z' + 8z = 0 if z = e-2x (sin 2x + cos 2x). Z = e-2x (sin 2x + cos 2x) Let u = 2x, then du = 2dx du = 2 dx Z = e-"(sin u + cos u) z'%3Dе"Dx(sin и + cos u) + (sin и + cos u)Dx(е ") z'%3Dе " (сos u — sin u) — е (sinu + cosu) coS и — е "и sin u — e"u sin u — e"и cos u dy = -2e¬u sin u du dy = -2e¬u sin u, du du Substitute = 2 using the Chain Rule, dx dz dy du (-2e¬" sin u)2 = -4e¬u sin u du du dx Substitute u = 2x, z' = -4e-2x sin 2x To get the second derivative, again let u = 2x then du = 2dx du = 2 dx z' = -4e-u sin u -4[e-"Dx(sin u) + sin u Dx(e¬u) 3 -4е И cos и — 4е -И sin u z" = dy — — 4е и соs и — 4e-U sin u du dy = -4e-u du Substitute cOS и — 4e"u sin u, = 2 using the Chain Rule, dx du dz dy du (-4е " cos u — 4е И sin u)2 3D —8е И сosu — 8e u sin u du du dx
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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Question
Please check the errors in the computation below. Insert the answer in the following table.
![Substitute u = 2x,
z" 3D —8е2х cos 2x — 8е 2x sin 2х
Show z" + 4z' + 8z = 0.
(-4e-2x sin 2x ) + 4(-8e-2x cos 2x – 8e-2x sin 2x) + 8[e¬2x (sin 2x + cos 2x)] = 0
-4e-2x sin 2x – 32e-2x cos 2x – 32e-2x sin 2x + 8e-2x sin 2x + 8e-2x cos 2x = 0
-24e
-2x
cos 2x — 28е-
e-2x sin 2x 0
IDENTIFIED ERROR
CORRECTION OF ERROR
EXPLANATION OF
CORRECTION](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc945876b-19fa-4726-848f-a10d3b966f75%2F76d896b1-6694-452d-83a3-f427ded68888%2Fj9qbgin_processed.png&w=3840&q=75)
Transcribed Image Text:Substitute u = 2x,
z" 3D —8е2х cos 2x — 8е 2x sin 2х
Show z" + 4z' + 8z = 0.
(-4e-2x sin 2x ) + 4(-8e-2x cos 2x – 8e-2x sin 2x) + 8[e¬2x (sin 2x + cos 2x)] = 0
-4e-2x sin 2x – 32e-2x cos 2x – 32e-2x sin 2x + 8e-2x sin 2x + 8e-2x cos 2x = 0
-24e
-2x
cos 2x — 28е-
e-2x sin 2x 0
IDENTIFIED ERROR
CORRECTION OF ERROR
EXPLANATION OF
CORRECTION
Transcribed Image Text:Use your knowledge on Higher-order derivative and Chain Rule to show that
z" + 4z' + 8z = 0 if z = e-2x (sin 2x + cos 2x).
Z = e-2x (sin 2x + cos 2x)
Let u = 2x, then
du = 2dx
du
= 2
dx
Z = e-"(sin u + cos u)
z'%3Dе"Dx(sin и + cos u) + (sin и + cos u)Dx(е ")
z'%3Dе " (сos u — sin u) — е (sinu + cosu)
coS и — е "и sin u — e"u sin u — e"и cos u
dy
= -2e¬u sin u
du
dy
= -2e¬u sin u,
du
du
Substitute
= 2 using the Chain Rule,
dx
dz
dy du
(-2e¬" sin u)2 = -4e¬u sin u
du
du dx
Substitute u = 2x,
z' = -4e-2x sin 2x
To get the second derivative, again let u = 2x then
du = 2dx
du
= 2
dx
z' = -4e-u sin u
-4[e-"Dx(sin u) + sin u Dx(e¬u)
3 -4е И cos и — 4е -И sin u
z" =
dy
— — 4е и соs и — 4e-U sin u
du
dy
= -4e-u
du
Substitute
cOS и — 4e"u sin u,
= 2 using the Chain Rule,
dx
du
dz
dy du
(-4е " cos u — 4е И sin u)2 3D —8е И сosu — 8e u sin u
du
du dx
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