Chapter 12: Q 72. (page 965)

Prove that
$\nabla (\frac{f (x, y)}{g (x, y)}) = \frac{g (x, y) \nabla f (x, y) - f (x, y) \nabla g (x, y)}{(g (x, y))^{2}}$
where $g (x, y) \neq 0$

Short Answer

Expert verified

Solve for $\nabla (\frac{f (x, y)}{g (x, y)}) = (i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y}) (\frac{f (x, y)}{g (x, y)})$ to prove the above relation.

Step by step solution

Given Information

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

We need to show that $\nabla (\frac{f (x, y)}{g (x, y)}) = \frac{g (x, y) \nabla f (x, y) - f (x, y) \nabla g (x, y)}{(g (x, y))^{2}}$

Use i∂∂x+j∂∂y for ∇

Substituting we get

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

$= i \frac{\partial}{\partial x} (\frac{f (x, y)}{g (x, y)}) + j \frac{\partial}{\partial y} (\frac{f (x, y)}{g (x, y)})$

Simplification

\nabla (\frac{f (x, y)}{g (x, y)}) = i [\frac{g (x, y) \frac{\partial}{\partial x} (f (x, y)) - f (x, y) \frac{\partial}{\partial x} (g (x, y))}{[g (x, y)]^{2}}]

$+ j [\frac{g (x, y) \frac{\partial}{\partial y} (f (x, y)) - f (x, y) \frac{\partial}{\partial y} (g (x, y))}{[g (x, y)]^{2}}]$

Rearranging, we get

$\nabla (\frac{f (x, y)}{g (x, y)}) = \frac{g (x, y) i \frac{\partial}{\partial x} (f (x, y)) + g (x, y) j \frac{\partial}{\partial y} (f (x, y))}{[g (x, y)]^{2}}$

$+ \frac{f (x, y) i \frac{\partial}{\partial x} (g (x, y)) + f (x, y) j \frac{\partial}{\partial y} (g (x, y))}{[g (x, y)]^{2}}$

$\nabla (\frac{f (x, y)}{g (x, y)}) = \frac{g (x, y) (i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y}) (f (x, y))}{[g (x, y)]^{2}} + \frac{f (x, y) (i \frac{\partial}{\partial x} + j \frac{\partial}{\partial y}) (g (x, y))}{[g (x, y)]^{2}}$

$\nabla (\frac{f (x, y)}{g (x, y)}) = \frac{g (x, y) \cdot \nabla (f (x, y)) + f (x, y) \cdot \nabla (g (x, y))}{[g (x, y)]^{2}}$

$\nabla (\frac{f (x, y)}{g (x, y)}) = \frac{g (x, y) \nabla f (x, y) - f (x, y) \nabla g (x, y)}{(g (x, y))^{2}}$

Hence proved.

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Prove that
$\nabla (\frac{f (x, y)}{g (x, y)}) = \frac{g (x, y) \nabla f (x, y) - f (x, y) \nabla g (x, y)}{(g (x, y))^{2}}$
where $g (x, y) \neq 0$

Short Answer

Step by step solution

Given Information

Use i∂∂x+j∂∂y for ∇

Simplification

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Prove that∇f(x,y)g(x,y)=g(x,y)∇f(x,y)-f(x,y)∇g(x,y)(g(x,y))2whereg(x,y)≠0

Short Answer

Step by step solution

Given Information

Use i∂∂x+j∂∂y for ∇

Simplification

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Pure Maths

Geometry

Applied Mathematics

Probability and Statistics

Mechanics Maths

Statistics

Study anywhere. Anytime. Across all devices.

Prove that
$\nabla (\frac{f (x, y)}{g (x, y)}) = \frac{g (x, y) \nabla f (x, y) - f (x, y) \nabla g (x, y)}{(g (x, y))^{2}}$
where $g (x, y) \neq 0$