Leibnitz's theorem and L'Hospital's ruleEdexcel A-Level Further Maths: Revision notes

Section 1

Leibnitz's theorem

Leibnitz's theorem gives the $n$ th derivative of a product $y=uv$ : $\frac{\mathrm{d}^n}{\mathrm{d}x^n}(uv)=\sum_{r=0}^{n}\binom nr u^{(r)}v^{(n-r)}=uv^{(n)}+nu'v^{(n-1)}+\binom n2u''v^{(n-2)}+\dots+u^{(n)}v.$ The coefficients are binomial coefficients, because the pattern mirrors the expansion of $(a+b)^n$ . For $n=2$ this gives $u''v+2u'v'+uv''$ and for $n=3$ , $u'''v+3u''v'+3u'v''+uv'''$ . Choose $u$ to be the factor that becomes zero after a few differentiations (such as a polynomial), so most terms vanish.

Key termsLeibnitz's theorembinomial coefficient

Common mistake

Mixing up the order of derivatives. The $r$ th derivative of $u$ is paired with the $(n-r)$ th derivative of $v$ , and the coefficient is $\binom nr$ .

Section 2

Worked example with Leibnitz's theorem

Let $y=x^2\mathrm{e}^{3x}$ with $u=x^2$ ( $u'=2x$ , $u''=2$ , $u'''=0$ ) and $v=\mathrm{e}^{3x}$ ( $v^{(k)}=3^k\mathrm{e}^{3x}$ ). Then $y''=u''v+2u'v'+uv''=\mathrm{e}^{3x}(9x^2+12x+2).$ For any $n\ge2$ only three terms survive: $y^{(n)}=\mathrm{e}^{3x}\left(3^nx^2+2n3^{n-1}x+n(n-1)3^{n-2}\right).$ At $x=0$ , $y^{(4)}(0)=\binom42(2)(9)=108$ . A pattern valid for all $n$ can be found whenever one factor is a polynomial.

Key termssurviving terms

Exam tip

Write out $u,u',u'',\dots$ and $v,v',v'',\dots$ in two columns before combining.

Section 3

L'Hospital's rule

L'Hospital's rule: if $f(x)\to0$ and $g(x)\to0$ , or both tend to $\pm\infty$ , as $x\to a$ , then $\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$ provided the right-hand limit exists. The rule applies to indeterminate forms $\frac00$ and $\frac\infty\infty$ . Example: $\lim_{x\to0}\frac{\sin3x}{\sin5x}$ is $\frac00$ ; differentiating gives $\frac{3\cos3x}{5\cos5x}\to\frac35$ . Differentiate the numerator and the denominator separately; do not use the quotient rule.

Key termsL'Hospital's ruleindeterminate form

Common mistake

Using the quotient rule on $\frac fg$ . L'Hospital's rule differentiates top and bottom separately.

Section 4

Repeated applications

If the new quotient is still indeterminate, apply the rule again. Example: $\lim_{x\to0}\frac{2\sin x-\sin2x}{x-\sin x}$ . Differentiate: $\frac{2\cos x-2\cos2x}{1-\cos x}$ , still $\frac00$ . Again: $\frac{-2\sin x+4\sin2x}{\sin x}$ , still $\frac00$ . Again: $\frac{-2\cos x+8\cos2x}{\cos x}\to\frac{-2+8}{1}=6$ . You may simplify between steps, for example $\frac{-2\sin x+4\sin2x}{\sin x}=-2+8\cos x$ , which gives $6$ directly. Check the form at every step.

Key termsrepeated application

Exam tip

Test the form ( $\frac00$ or $\frac\infty\infty$ ) at every step, and stop as soon as a limit can be read off by substitution.

Section 5

Other indeterminate forms and when the rule fails

Forms such as $1^\infty$ , $0\times\infty$ are rearranged first. For $\lim_{x\to\infty}\left(1+\frac ax\right)^x$ put $y=\left(1+\frac ax\right)^x$ , so $\ln y=\frac{\ln(1+a/x)}{1/x}$ , which is $\frac00$ . L'Hospital gives $\frac{a}{1+a/x}\to a$ , so $y\to\mathrm{e}^a$ . For $\lim\frac{\mathrm{e}^x}{x^2}$ as $x\to\infty$ (form $\frac\infty\infty$ ) apply the rule twice to get $\frac{\mathrm{e}^x}{2}\to\infty$ . The rule must not be used if the form is not indeterminate: $\frac{1+\cos x}{x^2}$ tends to $\frac{2}{0^+}=+\infty$ , but a blind application gives $-\frac12$ , which is wrong.

Key termslogarithm trick

Common mistake

Applying the rule when the numerator tends to a non-zero constant. Substitute $x=a$ first.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Leibnitz's theorem and L'Hospital's rule

Let $y=x^2\mathrm{e}^{3x}$ .
Use Leibnitz's theorem to find the value of $\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4}$ at $x=0$ .2 marks
L'Hospital's rule states that if $f(x)$ and $g(x)$ both tend to $0$ , or both tend to $\pm\infty$ , as $x\to a$ , then $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$ , provided the second limit exists.
A student evaluates $\displaystyle\lim_{x\to0}\frac{1+\cos x}{x^2}$ as follows: “This is $\frac00$ . Differentiating gives $\frac{-\sin x}{2x}$ , which is again $\frac00$ . Differentiating again gives $\frac{-\cos x}{2}\to-\frac12$ .” Explain the error in the student's working and describe the true behaviour of the quotient as $x\to0$ .2 marks
Let $f(x)=\dfrac{2\sin x-\sin2x}{x-\sin x}$ for $x\neq0$ .
Show that applying L'Hospital's rule once to $\displaystyle\lim_{x\to0}f(x)$ gives an expression that is again of the form $\frac00$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).