Q 70. - 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

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

Step 1: Given Information

Considering function

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

We need to show that

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

Step 2: Simplification

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

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

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

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

Hence proved.