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Chapter 12: Q 68. (page 965)

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

Short Answer

Expert verified

Solve for $\nabla (α f) = i \frac{\partial (α f)}{\partial x} + j \frac{\partial (α f)}{\partial y}$ to prove this.

Step by step solution

01

Given Information

The function is

$\nabla (α f) (x, y)$

We need to show that $\nabla (α f) (x, y) = α \nabla f (x, y)$ .

02

Simplification

Simplifying, we get

$\nabla (α f) = i \frac{\partial (α f)}{\partial x} + j \frac{\partial (α f)}{\partial y}$

$\nabla (α f) = i \frac{α \partial f}{\partial x} + j \frac{α \partial f}{\partial y}$

$\nabla (α f) = α (i \frac{\partial f}{\partial x} + j \frac{\partial f}{\partial y})$

$\nabla (α f) = α \nabla f$

Hence

$\nabla (α f) = α \nabla f$

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