Find all first partial derivatives. z = ln(x^2 + y^2 + 1)

CheemnCatelvew 2021-02-21 Answered
Find all first partial derivatives. z=ln(x2+y2+1)
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Expert Answer

insonsipthinye
Answered 2021-02-22 Author has 83 answers
Step 1
Given
The equation is z=ln(x2+y2+1).
Step 2
To find all first partial derivatives .
To find dzdx.
dzdx(ln(x2+y2+1))
=1x2+y2+1ddx(x+y2+1)
=1x2+y2+1(2x)
dzdx=2xx2+y2+1
To find dzdy.
dzdy=ddy(ln(x2+y2+1))
=1x2+y2+1ddy(x2+y2+1)
=1x2+y2+1(2y)
dzdy=2yx2+y2+1
Therefore , The all partial derivatives are dzdx=2xx2+y2+1anddzdy=2yx2+y2+1.
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Jeffrey Jordon
Answered 2021-11-17 Author has 2064 answers

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