Find the derivative of the given function. Fill the missing expressions. Use logarithmic differentiation. y = x• secx Vx² + 1 Step 1. Take the absolute value of both sides. lyl = x• secx · /x2 + 1 Step 2. Take the natural logarithm of both sides. In ]y| = In x· secx /x² + 1 secx · Step 3. Expand using the properties of Logarithm Inly| = Inx + In|secx| + Step 4. Differentiate both sides. 1 D,(Inlyl) = D, [Inx + In|secx| + In(x² + 1)] 1 1 .· Dz(secx) + 1 D,(x² + 1) 2 x2 + 1 1 = -+ 1 secx tanx + 1 2x 2 x2 + 1 Step 5. Multiply both sides of the equation by y to solve for y' ' = y · (- tanx + (x2 + 1)) Step 6. Substitute y with the given function. y' 1 + 5cotx + 3(x +1) (1 – x2),

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...
icon
Related questions
Question
EVALUATION
Find the derivative of the given function. Fill the missing expressions. Use logarithmic
differentiation.
y = x· secx · Vx² + 1
Step 1. Take the absolute value of both sides.
lyl = |x - secx · Jx2 + 1
Step 2. Take the natural logarithm of both sides.
In ]y| = In x• secx · x2 + 1
Step 3. Expand using the properties of Logarithm
Inlyl = Inx + In|secx| +
Step 4. Differentiate both sides.
D,(In\y\) = Dx Inx + In|secx| +In(x² + 1)
1
1
= -+
.· Dz(secx) +
1
Dz(x² + 1)
2 x2 + 1
1
1
· secx tanx +:
1
2x
2 x2 + 1
Step 5. Multiply both sides of the equation by y to solve for y'
+ tanx +
(x² + 1)
Step 6. Substitute y with the given function.
1
y' =
+ 5cotx +
\3(x+ 1)
(1 – x²).
Transcribed Image Text:EVALUATION Find the derivative of the given function. Fill the missing expressions. Use logarithmic differentiation. y = x· secx · Vx² + 1 Step 1. Take the absolute value of both sides. lyl = |x - secx · Jx2 + 1 Step 2. Take the natural logarithm of both sides. In ]y| = In x• secx · x2 + 1 Step 3. Expand using the properties of Logarithm Inlyl = Inx + In|secx| + Step 4. Differentiate both sides. D,(In\y\) = Dx Inx + In|secx| +In(x² + 1) 1 1 = -+ .· Dz(secx) + 1 Dz(x² + 1) 2 x2 + 1 1 1 · secx tanx +: 1 2x 2 x2 + 1 Step 5. Multiply both sides of the equation by y to solve for y' + tanx + (x² + 1) Step 6. Substitute y with the given function. 1 y' = + 5cotx + \3(x+ 1) (1 – x²).
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps

Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning