4. What is the correct expression of the directional derivative for p = xy ln x + sec-[sinh(xy)]? y coth(xy) х coth(ху) \dy do A. ds dx +(x In x + y + y lnx + ds /sinh²(xy) – 1) ds у coth (ху) /sinh²(xy) – 1, sinh²(xy) – 1/ х coth(ху) dy Vsinh?(xy) – 1 В. dФ y + y ln x + dx +(x In x + dx dp C. = [y + y ln x + y csch(xy)]. ds dy + [x In x ± x csch(xy)] 75 ds D. do = [y + y ln x ± y csch(xy)]dx + [x ln x ± x csch(xy)]d

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Question
4.
What is the correct expression of the directional derivative for o = xy lIn x + sec-[sinh(xy)]?
do
A.
ds
y coth(xy)
Vsinh?(xy) – 1) ds
y coth(xy)
х coth(ху)
dy
dx
+(x In x +
y + y ln x +
Vsinh?(xy) – 1/) ds
х coth(ху)
B. dọ = (y +yln x +
/sinh?(хy) — 1)
dx
dx +(x In x +
|dy
sinh²(xy) – ,
do
= [y + y lnx + y csch(xy)]
ds
dy
+ [x In x ±x csch(xy)] 5
C.
D. do = [y + y ln x+y csch(xy)]dx + [x ln x +x csch(xy)]d
Transcribed Image Text:4. What is the correct expression of the directional derivative for o = xy lIn x + sec-[sinh(xy)]? do A. ds y coth(xy) Vsinh?(xy) – 1) ds y coth(xy) х coth(ху) dy dx +(x In x + y + y ln x + Vsinh?(xy) – 1/) ds х coth(ху) B. dọ = (y +yln x + /sinh?(хy) — 1) dx dx +(x In x + |dy sinh²(xy) – , do = [y + y lnx + y csch(xy)] ds dy + [x In x ±x csch(xy)] 5 C. D. do = [y + y ln x+y csch(xy)]dx + [x ln x +x csch(xy)]d
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