Chapter 12: Q 70. (page 965)

Prove that
$\nabla (f (x, y) g (x, y)) = f (x, y) \nabla g (x, y) + g (x, y) \nabla f (x, y)$

Short Answer

Expert verified

Solve $\nabla (α f (x, y) + β g (x, y)) = i \frac{\partial (α f (x, y) + β g (x, y))}{\partial x} + j \frac{\partial (α f (x, y) + β g (x, y))}{\partial y}$ for proving above relation.

Step by step solution

Given Information

Considering function

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

We need to show that

$\nabla (α f (x, y) + β g (x, y)) = α \nabla f (x, y) + β \nabla g (x, y)$

Simplification

Solving for $\nabla (α f (x, y) + β g (x, y)) = i \frac{\partial (α f (x, y) + β g (x, y))}{\partial x} + j \frac{\partial (α f (x, y) + β g (x, y))}{\partial y}$

$\nabla (α f (x, y) + β g (x, y)) = i [\frac{α \partial (f (x, y)) + β \partial (g (x, y))}{\partial x}] + j [\frac{α \partial (f (x, y)) + β \partial (g (x, y))}{\partial y}]$

$\nabla (f (x, y) + g (x, y)) = [α i \frac{\partial (f (x, y))}{\partial x} + α j \frac{\partial (f (x, y))}{\partial y}] + [β i \frac{\partial (g (x, y))}{\partial x} + β j \frac{\partial (g (x, y))}{\partial y}]$

$\nabla (α f (x, y) + β g (x, y)) = α \nabla f (x, y) + β \nabla g (x, y)$

Hence proved.

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Prove that
$\nabla (f (x, y) g (x, y)) = f (x, y) \nabla g (x, y) + g (x, y) \nabla f (x, y)$

Short Answer

Step by step solution

Given Information

Simplification

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

Short Answer

Step by step solution

Given Information

Simplification

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Logic and Functions

Geometry

Theoretical and Mathematical Physics

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Statistics

Study anywhere. Anytime. Across all devices.

Prove that
$\nabla (f (x, y) g (x, y)) = f (x, y) \nabla g (x, y) + g (x, y) \nabla f (x, y)$