Partial derivatives: Find all first- and second-order partial derivatives for the following functions:

f(x,y,z)=ln(x+y+z)

Short Answer

Expert verified

2fx2=1(x+y+z)2

2fy2=1(x+y+z)2

2fz2=1(x+y+z)2

Step by step solution

01

Step 1. Given information

f(x,y,z)=ln(x+y+z)

02

Step 2. Differentiate

Partially differentiating with respect to x,y and z,

fx=1x+y+zfy=1x+y+zfz=1x+y+z

Again differentiating with respect to x, y and z,

2fx2=1(x+y+z)22fzx=1(x+y+z)2=2fxz2fyx=1(x+y+z)2=2fxy

Again differentiating with respect to y and z,

2fy2=1(x+y+z)22fzy=1(x+y+z)2=2fyz

Now,

2fz2=1(x+y+z)2

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.

Sign-up for free