Answer
See proof
Work Step by Step
Let's start with the left side of the equation to be proved: \[ \begin{align*} \text{div}(k\mathbf{F}) &= \frac{\partial(kM)}{\partial x} + \frac{\partial(kN)}{\partial y} + \frac{\partial(kP)}{\partial z} \\ &= k\cdot\frac{\partial M}{\partial x} + k\cdot\frac{\partial N}{\partial y} + k\cdot\frac{\partial P}{\partial z} \\ &= k\cdot\left(\frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} + \frac{\partial P}{\partial z}\right) \\ &= k\cdot\text{div}(\mathbf{F}) \end{align*} \] Hence, proved. Note: (1): Assume that \(\mathbf{F} = \langle M, N, P \rangle\), so that \(k\mathbf{F} = \langle kM, kN, kP \rangle\). (2): Use the constant multiple rule for differentiation. (3): Take out \(k\) as the common factor. (4): The expression inside the bracket is the definition of \(\text{div}(\mathbf{F})\).
