Q 72. - Calculus

Question

Prove that

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

where $g(x,y)\ne 0$

Step-by-Step Solution

Answer

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

Step 1: Given Information

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

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

Step 2: Use i ∂ ∂ x + j ∂ ∂ y for ∇

Substituting we get

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

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

Step 3: Simplification

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

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

Rearranging, we get

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

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

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

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

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

Hence proved.