Core Pure · Series · Year 2

Higher Derivatives

Differentiating more than once or twice is nothing new in principle — just repeated application of rules you already know — but keeping track of the pattern as the powers climb is the real skill, and it's exactly what the next lesson on Maclaurin series depends on.

8 min read 2 worked examples

Higher derivatives

Notation — the nth derivative

You already differentiate once to get $\dfrac{dy}{dx}=f'(x)$, and twice to get $\dfrac{d^2y}{dx^2}=f''(x)$. Differentiating $n$ times in total gives the $n$th derivative, written $\dfrac{d^ny}{dx^n}=f^{(n)}(x)$. There's nothing conceptually new here — just repeated application of the differentiation rules you already know (chain rule, product rule, quotient rule), keeping careful track of how the expression changes shape each time.

Example 1 · Finding a third derivative

Given $y=\ln(1+2x)$, find $\dfrac{d^3y}{dx^3}$, and evaluate it when $x=0$.

$\dfrac{dy}{dx}=\dfrac2{1+2x}=2(1+2x)^{-1}$.

$\dfrac{d^2y}{dx^2}=2\times(-1)(1+2x)^{-2}\times2=-4(1+2x)^{-2}$.

$\dfrac{d^3y}{dx^3}=-4\times(-2)(1+2x)^{-3}\times2=16(1+2x)^{-3}$.

At $x=0$: $\dfrac{d^3y}{dx^3}=16(1)^{-3}=16$.

Spotting the pattern

For functions built from $\ln$, $e^x$ or simple powers, successive derivatives often follow a clean pattern in their coefficients, which you can sometimes guess and then confirm, rather than differentiating from scratch every time. This becomes essential in the next lesson, where you'll need $f^{(r)}(0)$ for arbitrarily large $r$.

Example 2 · Using implicit differentiation to find a recurrence

Given $f(x)=e^{3x^2}$, show that $f'(x)=6xf(x)$, and hence find $f''(x)$ and $f'''(x)$ in terms of $x$ and $f(x)$.

By the chain rule, $f'(x)=e^{3x^2}\times6x=6xe^{3x^2}=6xf(x)$, as required.

Differentiating $f'(x)=6xf(x)$ using the product rule: $f''(x)=6f(x)+6xf'(x)=6f(x)+6x(6xf(x))=6f(x)+36x^2f(x)=(6+36x^2)f(x)$.

Differentiating again: $f'''(x)=72xf(x)+(6+36x^2)f'(x)=72xf(x)+(6+36x^2)(6xf(x))=\big(72x+36x+216x^3\big)f(x)=\big(108x+216x^3\big)f(x)$.

Exam-style questions

Question 1 6 marks

Given $y=\cos2x$, find $\dfrac{d^4y}{dx^4}$ and evaluate it when $x=\dfrac\pi3$.

$\dfrac{dy}{dx}=-2\sin2x$; $\dfrac{d^2y}{dx^2}=-4\cos2x$; $\dfrac{d^3y}{dx^3}=8\sin2x$; $\dfrac{d^4y}{dx^4}=16\cos2x$. (M1, A1, A1, A1)

At $x=\dfrac\pi3$: $16\cos\dfrac{2\pi}3=16\times\left(-\dfrac12\right)=-8$. (M1, A1)

Question 2 5 marks

Given $y=x^2e^x$, find $\dfrac{d^3y}{dx^3}$.

$\dfrac{dy}{dx}=2xe^x+x^2e^x=(x^2+2x)e^x$ (M1, A1)

$\dfrac{d^2y}{dx^2}=(2x+2)e^x+(x^2+2x)e^x=(x^2+4x+2)e^x$ (M1, A1)

$\dfrac{d^3y}{dx^3}=(2x+4)e^x+(x^2+4x+2)e^x=(x^2+6x+6)e^x$ (A1)

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Maclaurin Series