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91Ó°ÊÓ

Find the first-order partial derivatives for the functions in Exercises 27–36.

fx,y,z=xy2x+z

Short Answer

Expert verified

The first-order partial derivatives are∂f∂x=zy2x+z2,∂f∂y=2xyx+z,∂f∂z=-xy2x+z2.

Step by step solution

01

Given 

The given function is: fx,y,z=xy2x+z

02

To find  

We have to find the first-order partial derivatives.

03

Calculation 

fx,y,z=xy2x+z∂f∂x=∂∂xxy2x+z-xy2∂∂xx+zx+z2∂f∂x=y2x+z-xy21x+z2∂f∂x=xy2+zy2-xy2x+z2∂f∂x=zy2x+z2fx,y,z=xy2x+z∂f∂y=∂∂yxy2x+z∂f∂y=2xyx+zfx,y,z=xy2x+z∂f∂z=∂∂zxy2x+z∂f∂z=-xy2x+z2∂∂zx+z∂f∂z=-xy2x+z21∂f∂z=-xy2x+z2Hence,∂f∂x=zy2x+z2,∂f∂y=2xyx+z,∂f∂z=-xy2x+z2.

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