FP1 2.3 - Leibniz's theorem | Edexcel A-Level Further Mathematics

Differentiate a product many times using binomial coefficients instead of repeatedly expanding the product rule. You will calculate higher derivatives, exploit polynomial factors that eventually disappear and use a general derivative to solve a high-order condition.

The product rule creates binomial coefficients

Can you find where the twentieth derivative of $x^2\mathrm{e}^{2x}$ is zero without differentiating twenty times? The eventual equation is only quadratic. The reason is that the polynomial factor stops contributing new derivatives after its second derivative.

You already know

The product rule is $(uv)'=u'v+uv'$ . From A-level Mathematics, $\binom nk=n!/(k!(n-k)!)$ , with the coefficients of $(a+b)^n$ given by the corresponding row of Pascal's triangle. Higher derivative notation $u^{(k)}$ means derivative order $k$ , not a power.

Differentiate the product rule again and collect identical terms:

\begin{aligned} (uv)''&=u''v+u'v'+u'v'+uv'' \\ &=u''v+2u'v'+uv'' \end{aligned}

One more differentiation gives $u'''v+3u''v'+3u'v''+uv'''$ . The coefficients are $1,2,1$ and $1,3,3,1$ , just as in binomial expansions.

In $n$ differentiations, choose which $k$ of the differentiation steps act on $u$ ; the other $n-k$ act on $v$ . There are $\binom nk$ such choices, each giving the same final product $u^{(k)}v^{(n-k)}$ . That counting argument explains the theorem.

Leibniz's theorem

If $u$ and $v$ have the required derivatives through order $n$ ,

\frac{\mathrm{d}^n}{\mathrm{d}x^n}(uv) =\sum_{k=0}^{n}\binom nk u^{(k)}v^{(n-k)}

Here $n$ is a non-negative integer, $u^{(0)}=u$ and $v^{(0)}=v$ . In each term the derivative orders add to $n$ .

Not in the formula booklet: learn it.

You may list the terms in the opposite order, since $\binom nk=\binom n{n-k}$ . What matters is matching each coefficient with a pair of derivative orders whose sum is the requested total.

Key idea

Use one row of binomial coefficients and let the derivative order move from one factor to the other.

The total derivative order provides a quick diagnostic check.

Calculating a specified higher derivative

List the derivatives of each factor before inserting them into the theorem. This keeps chain-rule factors and trigonometric signs out of the coefficient bookkeeping.

For a fourth derivative, use coefficients $1,4,6,4,1$ . A common exponential factor can then be taken out before collecting sine and cosine terms.

Worked example 1: a fourth derivative that returns to the original shape

Question. Use Leibniz's theorem to find the fourth derivative of $y=\mathrm{e}^x\sin x$ .

Plan. Use the fourth-order binomial coefficients and the sine derivative cycle, then collect terms.

Every derivative of $\mathrm{e}^x$ is $\mathrm{e}^x$ ; the successive sine derivatives are $\sin x,\cos x,-\sin x,-\cos x,\sin x$ . Thus

\begin{aligned} y^{(4)} &=\mathrm{e}^x\bigl(\sin x+4\cos x-6\sin x-4\cos x+\sin x\bigr) && \text{Leibniz coefficients} \\ &=-4\mathrm{e}^x\sin x && \text{cosine terms cancel} \end{aligned}

Why this step? Writing the signed trigonometric derivatives before collecting makes the cancellation visible; the binomial coefficients themselves are all positive.

Answer. $y^{(4)}=-4\mathrm{e}^x\sin x$ .

Check. Directly, $y'=\mathrm{e}^x(\sin x+\cos x)$ , $y''=2\mathrm{e}^x\cos x$ , $y'''=2\mathrm{e}^x(\cos x-\sin x)$ , leading to the same fourth derivative. Four derivatives return to the original shape with a multiplier of $-4$ .

Change only the exponential rate from $1$ to $2$ . Each exponential derivative now contributes another factor of $2$ .

Worked example 2: tracking the exponential rate

Question. Use Leibniz's theorem to find $\mathrm{d}^4y/\mathrm{d}x^4$ for $y=\mathrm{e}^{2x}\sin x$ .

Plan. Insert the exponential derivatives $2^j\mathrm{e}^{2x}$ into the same fourth-order pattern.

\begin{aligned} y^{(4)} &=\mathrm{e}^{2x}\bigl(16\sin x+32\cos x-24\sin x-8\cos x+\sin x\bigr) && \text{distribute derivative orders} \\ &=\mathrm{e}^{2x}(-7\sin x+24\cos x) && \text{collect like functions} \end{aligned}

Why this step? The term with two derivatives on each factor is $6(4\mathrm{e}^{2x})(-\sin x)$ , giving $-24\mathrm{e}^{2x}\sin x$ . Both the binomial and chain-rule factors are needed.

Answer. $\displaystyle \frac{\mathrm{d}^4y}{\mathrm{d}x^4}=\mathrm{e}^{2x}(-7\sin x+24\cos x)$ .

Check. At $x=0$ , the five terms give $0+32-0-8+0=24$ , agreeing with the simplified expression.

The theorem and these product derivatives are not printed in the formula booklet. In a show-that question, state the derivative being found as well as the final expression; a bare bracket can leave the conclusion unclear.

Watch out

Wrong: use $1,4,4,4,1$ for the fourth derivative. Right: the middle coefficient is $\binom42=6$ . Check Pascal's row before combining it with the derivative values.

Change the exponential rate once more.

A polynomial makes the general derivative short

Choose the polynomial as the factor whose derivative order increases in the sum. Once its derivative is zero, every later term vanishes. A degree-two factor therefore leaves at most three terms, however large $n$ is.

The same observation connects higher derivatives to Taylor coefficients: after finding $y^{(n)}$ , evaluate it at the expansion centre and divide by $n!$ . First obtain a formula valid for the requested derivative order.

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FP1 2.3 - Leibniz's theorem

The product rule creates binomial coefficients

Calculating a specified higher derivative

A polynomial makes the general derivative short

Continue this lesson.

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