In one line
Leibnitz's theorem gives the \(n\)th derivative of a product: \((uv)_n=\sum_{r=0}^{n}{}^nC_r\,u_{n-r}v_r\), which looks exactly like the binomial expansion.
Explained simply
You know the product rule \((uv)'=u'v+uv'\). Differentiate again and you get \(u''v+2u'v'+uv''\): the coefficients 1, 2, 1 are from \((a+b)^2\). In general, the \(n\)th derivative of a product follows the binomial pattern \(1,n,\frac{n(n-1)}{2},\dots\). That is Leibnitz's theorem. The syllabus asks for it without proof.
Statement: if \(u\) and \(v\) have \(n\)th derivatives, \((uv)_n=u_nv+{}^nC_1u_{n-1}v_1+{}^nC_2u_{n-2}v_2+\dots+uv_n\).
The theorem is most useful when one factor is a polynomial. A polynomial of degree \(k\) has zero derivatives after the \(k\)th, so only \(k+1\) terms of the sum survive, however large \(n\) is. That is why the choice of which factor to call \(v\) matters so much.
How it works
- Choose \(v\) as the factor that becomes zero after a few derivatives (usually a polynomial like \(x^2\)).
- Choose \(u\) as the factor whose \(n\)th derivative you know (like \(e^{ax}\), \(\sin ax\), \(\log x\)).
- Write the expansion; stop when \(v_r=0\).
- Substitute each \(u_{n-r}\) from the standard formulas and simplify.
Key formula
\((uv)_n=\displaystyle\sum_{r=0}^{n}{}^nC_r\,u_{n-r}\,v_r\), where \(u_k=\dfrac{d^ku}{dx^k}\), \({}^nC_r=\dfrac{n!}{r!(n-r)!}\), \({}^nC_1=n\), \({}^nC_2=\dfrac{n(n-1)}2\).
Quick example
Find \(y_n\) for \(y=x^2e^{3x}\). Take \(u=e^{3x}\), \(v=x^2\). Then \(u_k=3^ke^{3x}\), \(v_1=2x\), \(v_2=2\), \(v_3=0\).
\(y_n=3^ne^{3x}x^2+n\cdot3^{n-1}e^{3x}(2x)+\dfrac{n(n-1)}2\cdot3^{n-2}e^{3x}(2)\)
\(=3^{n-2}e^{3x}\left[9x^2+6nx+n(n-1)\right]\).
Check \(n=1\): \(3^{-1}e^{3x}(9x^2+6x)=e^{3x}(3x^2+2x)\), the direct derivative. Correct.
Common mistakes
- Choosing the polynomial as \(u\), so the series never stops.
- Forgetting the binomial coefficients and writing \(u_nv+u_{n-1}v_1+\dots\).
- Index mismatch: pairing \(u_{n-1}\) with \(v_2\) instead of \(v_1\).
How MU asks it
3–5-mark parts: "Find the \(n\)th derivative of \(x^3\log x\)" or "of \(x^2\sin x\)". It is also the key step in the "prove the relation" questions (next topic). For full marks state the theorem, identify \(u\) and \(v\), list the derivatives of \(v\) until zero, and simplify the final expression.