Prove that if u,v,w,t are functions of x possessing first
Prove that if $u,v,w,t$ are functions of $x$ possessing first and second derivatives, then \[ \begin{vmatrix} u & v & w \\ \bigl(ut\bigr)' & \bigl(vt\bigr)' & \bigl(wt\bigr)' \\ \bigl(ut\bigr)'' & \bigl(vt\bigr)'' & \bigl(wt\bigr)'' \end{vmatrix} = t^{2}\begin{vmatrix} u & v & w \\ u' & v' & w' \\ u'' & v'' & w'' \end{vmatrix}. \].
Explanation of Method
Two properties of determinants do all the work.
Step 1: Expand the entries
By the product rule, applied to each of $ut$, $vt$, $wt$: $\bigl(ut\bigr)' = u't+ut', \qquad \bigl(ut\bigr)'' = u''t+2u't'+ut''$.
Writing $P = [u,v,w]$, $P' = [u',v',w']$, $P'' = [u'',v'',w'']$ for the three basic rows, the determinant's rows are
\[ R_{1} = P, \qquad R_{2} = tP'+t'P, \qquad R_{3} = tP''+2t'P'+t''P. \]
Step 2: The two facts to be used
Only these are needed: \[\textbf{(i) } \text{a determinant is linear in each row separately;} \qquad \textbf{(ii) } \text{it vanishes if two rows are equal.}\]
Step 3: Split the second row
\[ \Delta = \begin{vmatrix} P \\ tP'+t'P \\ R_{3} \end{vmatrix} = \begin{vmatrix} P \\ tP' \\ R_{3} \end{vmatrix} + t'\begin{vmatrix} P \\ P \\ R_{3} \end{vmatrix}. \]
The second determinant has two identical rows, so it is zero. Factoring $t$ out of the remaining second row: \[\Delta = t\begin{vmatrix} P \\ P' \\ R_{3} \end{vmatrix}. \tag{1}\]
Step 4: Split the third row
\[ \begin{vmatrix} P \\ P' \\ tP''+2t'P'+t''P \end{vmatrix} = \begin{vmatrix} P \\ P' \\ tP'' \end{vmatrix} + 2t'\begin{vmatrix} P \\ P' \\ P' \end{vmatrix} + t''\begin{vmatrix} P \\ P' \\ P \end{vmatrix}. \]
The middle determinant repeats $P'$ and the last repeats $P$; both vanish. Factoring $t$ out of the third row of what remains: \[\begin{vmatrix} P \\ P' \\ R_{3} \end{vmatrix} = t\begin{vmatrix} P \\ P' \\ P'' \end{vmatrix}. \tag{2}\]
Step 5: Combine
Substituting (2) into (1): \[\Delta = t\cdot t\begin{vmatrix} u & v & w \\ u' & v' & w' \\ u'' & v'' & w'' \end{vmatrix} = t^{2}\begin{vmatrix} u & v & w \\ u' & v' & w' \\ u'' & v'' & w'' \end{vmatrix}.\]
Step 6: Verification
\[\text{Test } u = 1,\ v = x,\ w = x^{2},\ t = x:\quad \text{right side} = x^{2}\begin{vmatrix} 1 & x & x^{2} \\ 0 & 1 & 2x \\ 0 & 0 & 2 \end{vmatrix} = 2x^{2}\]
\[\text{Left side: } \begin{vmatrix} 1 & x & x^{2} \\ 1 & 2x & 3x^{2} \\ 0 & 2 & 6x \end{vmatrix} = 1\bigl(12x^{2}-6x^{2}\bigr)-x\bigl(6x-0\bigr)+x^{2}\bigl(2-0\bigr) = 2x^{2} \;\;\checkmark\]
\[\text{The factor } t^{2} \text{ — not } t^{3} \text{ — because the first row was left untouched} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\begin{vmatrix} u & v & w \\ \bigl(ut\bigr)' & \bigl(vt\bigr)' & \bigl(wt\bigr)' \\ \bigl(ut\bigr)'' & \bigl(vt\bigr)'' & \bigl(wt\bigr)'' \end{vmatrix} = t^{2}\begin{vmatrix} u & v & w \\ u' & v' & w' \\ u'' & v'' & w'' \end{vmatrix}\,}\]
Remark
Remark. The determinant on the right is the Wronskian of $u,v,w$. The identity therefore says that multiplying three functions by a common factor $t$ multiplies their Wronskian by $t^{3}$ — here one power is missing because the top row keeps $u,v,w$ rather than $ut,vt,wt$. In particular the left-hand determinant vanishes exactly when the Wronskian does, so linear dependence is unaffected by a common factor.