Q 68. - Calculus

Question

Prove that

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

Step-by-Step Solution

Answer

Solve for $\nabla \left(\alpha f\right)=i\frac{\partial \left(\alpha f\right)}{\partial x}+j\frac{\partial \left(\alpha f\right)}{\partial y}$ to prove this.

Step 1: Given Information

The function is

$\nabla \left(\alpha f\right)(x,y)$

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

Step 2: Simplification

Simplifying, we get

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

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

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

$\nabla \left(\alpha f\right)=\alpha \nabla f$

Hence

$\nabla \left(\alpha f\right)=\alpha \nabla f$