Q 71. - Calculus

Question

Prove that

$\nabla \left(f\right(x,y\left)g\right(x,y\left)\right)=f(x,y)\nabla g(x,y)+g(x,y)\nabla f(x,y)$

Step-by-Step Solution

Answer

The above relation can be proved using the formula

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

Step 1: Given Information

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

We need to show that

$\nabla \left(f\right(x,y\left)g\right(x,y\left)\right)=f(x,y)\nabla g(x,y)+g(x,y)\nabla f(x,y)$

Step 2: Simplification

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

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

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

$\nabla \left(f\right(x,y\left)g\right(x,y\left)\right)=f(x,y)\nabla \left(g\right(x,y\left)\right)+g(x,y)\nabla \left(f\right(x,y\left)\right)$

Hence proved.