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91Ó°ÊÓ

Prove that

∇(f(x,y)g(x,y))=f(x,y)∇g(x,y)+g(x,y)∇f(x,y)

Short Answer

Expert verified

Solve ∇(αf(x,y)+βg(x,y))=i∂(αf(x,y)+βg(x,y))∂x+j∂(αf(x,y)+βg(x,y))∂yfor proving above relation.

Step by step solution

01

Given Information

Considering function

f(x,y),g(x,y)

We need to show that

∇(αf(x,y)+βg(x,y))=α∇f(x,y)+β∇g(x,y)

02

Simplification

Solving for ∇(αf(x,y)+βg(x,y))=i∂(αf(x,y)+βg(x,y))∂x+j∂(αf(x,y)+βg(x,y))∂y

∇(αf(x,y)+βg(x,y))=iα∂(f(x,y))+β∂(g(x,y))∂x+jα∂(f(x,y))+β∂(g(x,y))∂y

∇(f(x,y)+g(x,y))=αi∂(f(x,y))∂x+αj∂(f(x,y))∂y+βi∂(g(x,y))∂x+βj∂(g(x,y))∂y

∇(αf(x,y)+βg(x,y))=α∇f(x,y)+β∇g(x,y)

Hence proved.

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