Homework Help is Here – Start Your Trial Now!
Math
Calculus
2. y = tan (cosh x) (sin x)(sech x) secx (use logarithmic differentiation)

2. y = tan (cosh x) (sin x)(sech x) secx (use logarithmic differentiation)

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
Chapter2: Graphical And Tabular Analysis
Section2.1: Tables And Trends
Problem 1TU: If a coffee filter is dropped, its velocity after t seconds is given by v(t)=4(10.0003t) feet per...
See similar textbooks
icon
Related questions
Question
100%

[question#2] (Example solution is provided in the photo, kindly follow the format). 

 

 

Find dy. DO NOT SIMPLIFY YOUR ANSWERS. dx 1. log₂ 09²(√2x² + y) + tan(x² − π²) = sin^¹(x² − 1) + = - 2. y tan (cosh x) (sin x) (sech x)** (use logarithmic differentiation) secx
Transcribed Image Text:Find dy. DO NOT SIMPLIFY YOUR ANSWERS. dx 1. log₂ 09²(√2x² + y) + tan(x² − π²) = sin^¹(x² − 1) + = - 2. y tan (cosh x) (sin x) (sech x)** (use logarithmic differentiation) secx
fa) - 2u sinh (x-1) ta - X = Note: DOM F(x) E IR f'(x) = (2xanh (x-1) +2²-x) f'(x) = d (2xsinh (x-1) + = (2x) - 2 / (x) dy dx f'(x) = 2x coch (x-1) + 2² In (2) + 24h(x-1)- / ('(x) is continuous on (0,1), therefore the function is differentiable on (0,1) f(0) = 2 (0) 6inh (0-1) +2°-0 f(₁) = 2(1) sinh (1-1) +2'-1 Fli) - 2 sinh lo) t flo) = 0+1 -0 flo) = 1 there is no x value parallel to the line a=0 b=1 7 where the tangent line at xis that passes through the end points lim x + -2²¯ f(x)= x³-4x² (x+2) ² (-2) ³ -4 (-2) ² (-2+2) ² 8-4 (9) O = RHL since to re LAL CLONT asyMPTOTES: lim x +∞ x² +4x +4 llence y = x- 8 there fore there is a vertical J at x = -2 (f(x)), lim x++oo (f(x)-ax) x³-4y² +x X³-√x²-x - 8x² -Ux lim x + 2+ f(x)= x³-Yv² (x+2) f(x) = (-2)² -4 (-2)² (-2+2)2 O 8y² +324 32 2 8x +32 asymptote
Transcribed Image Text:fa) - 2u sinh (x-1) ta - X = Note: DOM F(x) E IR f'(x) = (2xanh (x-1) +2²-x) f'(x) = d (2xsinh (x-1) + = (2x) - 2 / (x) dy dx f'(x) = 2x coch (x-1) + 2² In (2) + 24h(x-1)- / ('(x) is continuous on (0,1), therefore the function is differentiable on (0,1) f(0) = 2 (0) 6inh (0-1) +2°-0 f(₁) = 2(1) sinh (1-1) +2'-1 Fli) - 2 sinh lo) t flo) = 0+1 -0 flo) = 1 there is no x value parallel to the line a=0 b=1 7 where the tangent line at xis that passes through the end points lim x + -2²¯ f(x)= x³-4x² (x+2) ² (-2) ³ -4 (-2) ² (-2+2) ² 8-4 (9) O = RHL since to re LAL CLONT asyMPTOTES: lim x +∞ x² +4x +4 llence y = x- 8 there fore there is a vertical J at x = -2 (f(x)), lim x++oo (f(x)-ax) x³-4y² +x X³-√x²-x - 8x² -Ux lim x + 2+ f(x)= x³-Yv² (x+2) f(x) = (-2)² -4 (-2)² (-2+2)2 O 8y² +324 32 2 8x +32 asymptote
Expert Solution
steps

Step by step

Solved in 3 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Functions and Change: A Modeling Approach to Coll…
Functions and Change: A Modeling Approach to Coll…
Algebra
ISBN:
9781337111348
Author:
Bruce Crauder, Benny Evans, Alan Noell
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS