Q 69, - Calculus

Question

Prove that

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

$\alpha ,\beta$ are constants.

Step-by-Step Solution

Answer

The relation can be proved using $\nabla \left(f\right(x,y)+g(x,y\left)\right)=i\frac{\partial \left(f\right(x,y)+g(x,y\left)\right)}{\partial x}+j\frac{\partial \left(f\right(x,y)+g(x,y\left)\right)}{\partial y}$

Step 1: Given Information

Let the function be

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

We need to show that $\nabla \left(f\right(x,y)+g(x,y\left)\right)=\nabla f(x,y)+\nabla g(x,y)$

Step 2: Simplification

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

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

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

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

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

Hence proved.