দেওয়া আছে, $$ \phi = 2x^3y^2z^4 $$ আমরা জানি, ভেক্টর ডিফারেনশিয়াল অপারেটর $\vec{\nabla}$ হলো, $$ \vec{\nabla} = \frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k} $$ প্রথমে $\vec{\nabla} \phi$ বা $\phi$-এর গ্র্যাডিয়েন্ট নির্ণয় করি: $$ \begin{aligned} \vec{\nabla} \phi &= \left( \frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k} \right) (2x^3y^2z^4) \\ &= \frac{\partial}{\partial x}(2x^3y^2z^4) \hat{i} + \frac{\partial}{\partial y}(2x^3y^2z^4) \hat{j} + \frac{\partial}{\partial z}(2x^3y^2z^4) \hat{k} \\ &= 6x^2y^2z^4 \hat{i} + 4x^3yz^4 \hat{j} + 8x^3y^2z^3 \hat{k} \end{aligned} $$ এখন, $\vec{\nabla} \cdot \vec{\nabla} \phi$ বা ল্যাপলাসিয়ান নির্ণয় করি: $$ \begin{aligned} \vec{\nabla} \cdot \vec{\nabla} \phi &= \left( \frac{\partial}{\partial x} \hat{i} + \frac{\partial}{\partial y} \hat{j} + \frac{\partial}{\partial z} \hat{k} \right) \cdot (6x^2y^2z^4 \hat{i} + 4x^3yz^4 \hat{j} + 8x^3y^2z^3 \hat{k}) \\ &= \frac{\partial}{\partial x}(6x^2y^2z^4) + \frac{\partial}{\partial y}(4x^3yz^4) + \frac{\partial}{\partial z}(8x^3y^2z^3) \\ &= 12xy^2z^4 + 4x^3z^4 + 24x^3y^2z^2 \end{aligned} $$ অতএব, নির্ণেয় মান, $$ \vec{\nabla} \cdot \vec{\nabla} \phi = 12xy^2z^4 + 4x^3z^4 + 24x^3y^2z^2 $$