In one line
Compute the needed partial derivatives, combine them as the question asks, simplify using algebraic identities, and only then apply the operator a second time.
Explained simply
These "prove that" questions look scary because of the long operator, but they are just careful bookkeeping. It is like cooking a dish in two rounds: finish the first round neatly (simplify) before starting the second. The secret is usually an algebraic identity that collapses a big fraction into a simple one.
Most of these questions are built so that a big expression simplifies to something tiny, such as \(\dfrac{3}{x+y+z}\). If your working is getting longer and longer instead of shorter, stop and look for a factorisation you have missed. Recognising the standard identities is half the marks.
Technically, the operator \(\left(\dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial y}+\dfrac{\partial}{\partial z}\right)u=u_x+u_y+u_z\). Its square means applying it twice, not squaring the answer: \(\left(\dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial y}+\dfrac{\partial}{\partial z}\right)^2u=\left(\dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial y}+\dfrac{\partial}{\partial z}\right)\left(u_x+u_y+u_z\right)\).
How it works
- Find \(u_x\), \(u_y\), \(u_z\) (use symmetry: if \(u\) is symmetric, the others follow by changing letters).
- Add them and simplify using identities such as \(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\).
- Apply the operator again to the simplified result.
- Collect terms to reach the required expression.
Key formula
Identity used most often: \(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\). Operator square: \(\left(\sum\dfrac{\partial}{\partial x}\right)^2u=\sum\dfrac{\partial}{\partial x}\left(\sum u_x\right)\).
Quick example
If \(u=\log(x^3+y^3+z^3-3xyz)\), prove \(\left(\dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial y}+\dfrac{\partial}{\partial z}\right)^2u=\dfrac{-9}{(x+y+z)^2}\).
Step 1: \(u_x=\dfrac{3x^2-3yz}{x^3+y^3+z^3-3xyz}\), and by symmetry \(u_y=\dfrac{3y^2-3zx}{(\dots)}\), \(u_z=\dfrac{3z^2-3xy}{(\dots)}\).
Step 2: \(u_x+u_y+u_z=\dfrac{3(x^2+y^2+z^2-xy-yz-zx)}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}=\dfrac{3}{x+y+z}\).
Step 3: Apply the operator again: \(\dfrac{\partial}{\partial x}\dfrac{3}{x+y+z}=\dfrac{-3}{(x+y+z)^2}\), the same for \(y\) and \(z\). Sum \(=\dfrac{-9}{(x+y+z)^2}\). Proved.
Common mistakes
- Squaring the first-order sum, i.e. writing \((u_x+u_y+u_z)^2\); the operator square means differentiate twice.
- Not simplifying in Step 2, then drowning in quotient-rule algebra.
- Writing \(u_y\) by symmetry without checking that the function really is symmetric.
How MU asks it
A favourite 5/6-mark question. Variants: "If \(u=\log(\tan x+\tan y+\tan z)\), prove \(\sum\sin2x\,u_x=2\)", "If \(u=e^{xyz}\), find \(u_{xyz}\)" (answer \(e^{xyz}(1+3xyz+x^2y^2z^2)\)). For full marks: show at least one derivative fully, state "by symmetry" for the rest only when valid, show the factorisation step, and finish with "Hence proved".