Differentiation: further techniques, higher derivatives, stationary points

Differentiation – Further Techniques, Higher Derivatives & Stationary Points (Cambridge Pure Mathematics 3)

1. Quick Reference – Differentiation Rules Required by Syllabus 3.4

Rule	Formula	Typical Use in A‑Level Questions
Power rule	\(\displaystyle \frac{d}{dx}x^{n}=nx^{\,n-1}\quad (n\in\mathbb{R})\)	Algebraic functions, simplifying higher‑order derivatives
Product rule	\(\displaystyle \frac{d}{dx}[u\,v]=u'\,v+u\,v'\)	Products such as \(x^{2}\sin x\), \(e^{x}\ln x\)
Quotient rule	\(\displaystyle \frac{d}{dx}\!\left(\frac{u}{v}\right)=\frac{u'v-u v'}{v^{2}}\)	Rational functions, e.g. \(\dfrac{e^{x}}{1+x^{2}}\)
Chain rule (composite functions)	\(\displaystyle \frac{d}{dx}\,f(g(x))=f'\bigl(g(x)\bigr)\,g'(x)\)	Any nested function – exponential, log, trig, inverse trig …
Exponential & logarithmic	\(\displaystyle \frac{d}{dx}a^{x}=a^{x}\ln a,\qquad \frac{d}{dx}e^{u}=e^{u}u',\qquad \frac{d}{dx}\ln u=\frac{u'}{u}\)	Growth/decay, curve sketching, optimisation
Trigonometric	\(\displaystyle \frac{d}{dx}\sin u=\cos u\,u',\; \frac{d}{dx}\cos u=-\sin u\,u',\; \frac{d}{dx}\tan u=\sec^{2}u\,u'\)	Oscillatory motion, periodic functions
Inverse trigonometric	\(\displaystyle \frac{d}{dx}\sin^{-1}u=\frac{u'}{\sqrt{1-u^{2}}},\; \frac{d}{dx}\cos^{-1}u=-\frac{u'}{\sqrt{1-u^{2}}},\; \frac{d}{dx}\tan^{-1}u=\frac{u'}{1+u^{2}}\)	Integration by substitution, geometry problems
Hyperbolic (optional for extension)	\(\displaystyle \frac{d}{dx}\sinh u=\cosh u\,u',\; \frac{d}{dx}\cosh u=\sinh u\,u'\)	Advanced mechanics, differential equations
Implicit differentiation	\(\displaystyle \text{Differentiate }F(x,y)=0\text{ w.r.t. }x\text{ and solve for }\frac{dy}{dx}\)	Circles, ellipses, curves defined by \(x^{2}y+e^{y}=3\)
Parametric differentiation	\(\displaystyle \frac{dy}{dx}= \frac{dy/dt}{dx/dt},\qquad \frac{d^{2}y}{dx^{2}}=\frac{d}{dt}\!\left(\frac{dy}{dx}\right)\Big/\frac{dx}{dt}\)	Projectile motion, cycloids, Lissajous curves
Inverse‑function rule	\(\displaystyle \frac{d}{dx}f^{-1}(x)=\frac{1}{f'\bigl(f^{-1}(x)\bigr)}\)	Finding \(\dfrac{dx}{dy}\) when \(y=f(x)\) is easier

2. Higher‑Order Derivatives

2.1 Notation

• \(f'(x)=\dfrac{dy}{dx}\) – first derivative
• \(f''(x)=\dfrac{d^{2}y}{dx^{2}}\) – second derivative
• \(f^{(n)}(x)=\dfrac{d^{n}y}{dx^{n}}\) – \(n^{\text{th}}\) derivative

2.2 Leibniz’s Rule (Product) – useful for \(n\ge2\)

\[ \frac{d^{n}}{dx^{n}}\,[u(x)v(x)]=\sum_{k=0}^{n}\binom{n}{k}\,u^{(k)}(x)\,v^{(n-k)}(x) \]

2.3 Example – Successive Differentiation of a Composite Product

Find the first three derivatives of \(f(x)=e^{3x}\sin 2x\).

\[ \begin{aligned} f'(x)&=e^{3x}\bigl(3\sin2x+2\cos2x\bigr),\\[4pt] f''(x)&=e^{3x}\bigl(13\sin2x+12\cos2x\bigr),\\[4pt] f^{(3)}(x)&=e^{3x}\bigl(55\sin2x+66\cos2x\bigr). \end{aligned} \]

2.4 Higher‑Order Differentiation of Parametric Curves

\[ \frac{d^{2}y}{dx^{2}}= \frac{d}{dt}\!\left(\frac{dy}{dx}\right)\Big/\frac{dx}{dt}, \qquad \frac{d^{3}y}{dx^{3}}= \frac{d}{dt}\!\left(\frac{d^{2}y}{dx^{2}}\right)\Big/\frac{dx}{dt}, \text{ etc.} \]

3. Stationary Points & Curve Sketching

3.1 Definition

A point \(x=a\) on the curve \(y=f(x)\) is stationary if \(f'(a)=0\) (the derivative must exist).

3.2 Classification – Second‑Derivative Test

1. If \(f'(a)=0\) and \(f''(a)>0\) → local minimum.
2. If \(f'(a)=0\) and \(f''(a)<0\) → local maximum.
3. If \(f'(a)=0\) and \(f''(a)=0\) → inconclusive – use the higher‑order test.

3.3 Higher‑Order Test (when the second derivative also vanishes)

Let \(m\ge3\) be the smallest integer such that \(f^{(m)}(a)eq0\) while all lower derivatives are zero.

• If \(m\) is even:
  • \(f^{(m)}(a)>0\) → local minimum.
  • \(f^{(m)}(a)<0\) → local maximum.
• If \(m\) is odd → \(x=a\) is a point of inflection (the curve crosses its tangent).

3.4 Points of Inflection

A point \(x=c\) is a point of inflection if the concavity changes sign there. Sufficient conditions:

• \(f''(c)=0\) and \(f'''(c)eq0\); or
• The sign of \(f''(x)\) is different on the two sides of \(c\).

3.5 Worked Example – Full Stationary‑Point Analysis (A‑Level style)

Find and classify all stationary points of

\[ f(x)=x^{4}-4x^{3}+6x^{2}+1. \]

1. First derivative: \[ f'(x)=4x^{3}-12x^{2}+12x=4x\bigl(x^{2}-3x+3\bigr). \]
2. Stationary points satisfy \(f'(x)=0\). The quadratic factor has discriminant \(\Delta=9-12<0\); therefore the only real solution is \(x=0\).
3. Second derivative: \[ f''(x)=12x^{2}-24x+12=12\,(x-1)^{2}\ge0\quad\forall x. \] Hence \(f''(0)=12>0\).
4. Classification: because \(f'(0)=0\) and \(f''(0)>0\), the point \((0,\,f(0))=(0,1)\) is a local minimum.
5. Inflection check: \[ f'''(x)=24x-24,\qquad f'''(1)=0. \] Since \(f''(x)=12(x-1)^{2}\) never changes sign, \(x=1\) is not an inflection point; it is a stationary point of horizontal tangent but not a local extremum.

3.6 Quick Sketch Checklist (Paper 1 & Paper 3)

• Find \(f'(x)=0\) → stationary points.
• Use the second‑derivative test; if \(f''=0\) apply the higher‑order test.
• Locate points where \(f''(x)=0\) → possible inflection points; confirm by a sign change of \(f''\) or by checking \(f'''(c)eq0\).
• Identify asymptotes (vertical from zeros of the denominator, horizontal/slant from limits).
• Plot a few representative points (including intercepts) to verify the overall shape.

4. Simple Polynomial & Rational Example (Pure Mathematics 1 – Paper 1)

This example mirrors the typical “curve‑sketch” question in Paper 1, using only algebraic functions.

Problem

Sketch the curve \(y=\displaystyle\frac{x^{3}-6x}{x^{2}+1}\). Find all stationary points, classify them, and indicate any points of inflection.

Solution

1. First derivative (quotient rule): \[ y'=\frac{(3x^{2}-6)(x^{2}+1)-(x^{3}-6x)(2x)}{(x^{2}+1)^{2}} =\frac{x^{4}-6x^{2}+6}{(x^{2}+1)^{2}}. \]
2. Stationary points: set the numerator to zero. \[ x^{4}-6x^{2}+6=0\;\Longrightarrow\;x^{2}=3\pm\sqrt{3}. \] Hence \[ x=\pm\sqrt{\,3+\sqrt3\,},\qquad x=\pm\sqrt{\,3-\sqrt3\,}. \] All four values are real.
3. Second derivative (differentiate \(y'\) using the quotient rule again, or differentiate the simplified form): \[ y''=\frac{4x\bigl(x^{4}-6x^{2}+6\bigr)-2x(x^{2}+1)(2x^{3}-12x)}{(x^{2}+1)^{3}} =\frac{4x^{5}-24x^{3}+24x}{(x^{2}+1)^{3}}. \] Evaluate \(y''\) at each stationary abscissa: \[ \begin{array}{c|c|c} x & y' =0 & y'' \\ \hline \sqrt{3+\sqrt3} & 0 & <0\ (\text{maximum})\\[2pt] -\sqrt{3+\sqrt3} & 0 & <0\ (\text{maximum})\\[2pt] \sqrt{3-\sqrt3} & 0 & >0\ (\text{minimum})\\[2pt] -\sqrt{3-\sqrt3} & 0 & >0\ (\text{minimum}) \end{array} \] (The sign follows from substituting the numerical values; algebraically one can factor \(y''=4x(x^{2}-3)(x^{2}+1)^{-3}\).)
4. Points of inflection: solve \(y''=0\): \[ 4x(x^{2}-3)=0\;\Longrightarrow\;x=0,\;x=\pm\sqrt3. \] Check the sign of \(y''\) on either side of each candidate:
   • \(x=-\sqrt3\): \(y''\) changes from positive (left) to negative (right) → inflection.
   • \(x=0\): sign changes from negative to positive → inflection.
   • \(x=+\sqrt3\): sign changes from negative to positive → inflection.
5. Other useful features:
   • Vertical asymptotes – none, because the denominator \(x^{2}+1>0\) for all real \(x\).
   • Horizontal asymptote – divide numerator and denominator by \(x^{2}\): \[ y=\frac{x-\dfrac{6}{x}}{1+\dfrac{1}{x^{2}}\;} \xrightarrow[x\to\pm\infty]{} x, \] so the curve has an oblique (slant) asymptote \(y=x\).
   • Intercepts: \(y=0\) when \(x^{3}-6x=0\Rightarrow x=0,\pm\sqrt6\); \(x=0\) gives \(y=0\). \(y\)-intercept is also \((0,0)\).
6. Sketch: Plot the four stationary points, the three inflection points, the slant asymptote \(y=x\), and a few extra points (e.g. \(x=\pm2\)). Connect the pieces respecting the sign of the first derivative on each interval.

5. Application‑Focused Techniques (Syllabus 3.4 – 3.6)

5.1 Implicit & Parametric Curves – Tangents & Curvature

• For an implicit curve \(F(x,y)=0\): \[ \frac{dy}{dx}= -\frac{F_{x}}{F_{y}}. \]
• Second derivative (useful for inflection): \[ \frac{d^{2}y}{dx^{2}}=-\frac{F_{xx}+2F_{xy}\frac{dy}{dx}+F_{yy}\left(\frac{dy}{dx}\right)^{2}}{F_{y}}. \]
• Parametric example – unit circle \(x=\cos t,\;y=\sin t\): \[ \frac{dy}{dx}= \frac{\cos t}{-\,\sin t}= -\cot t,\qquad \frac{d^{2}y}{dx^{2}}= \frac{1}{\sin^{3}t}. \]

5.2 Optimisation & Related‑Rates

Typical solution steps (AO1–AO3):

1. Translate the problem into an algebraic relationship between the quantities.
2. Differentiate (explicitly, implicitly or parametrically) to relate the required rates.
3. Set the derivative of the quantity to be optimised to zero → locate stationary points.
4. Confirm a maximum or minimum with the second‑derivative test or a sign chart.

5.3 Motion Along a Curve

If a particle moves on \(y=f(x)\) with horizontal speed \(\dfrac{dx}{dt}=v_{x}(t)\), then

\[ \frac{dy}{dt}=f'(x)\frac{dx}{dt},\qquad \frac{d^{2}y}{dt^{2}}=f''(x)\Bigl(\frac{dx}{dt}\Bigr)^{2}+f'(x)\frac{d^{2}x}{dt^{2}}. \]

These formulas appear in A‑Level mechanics questions involving trajectories or curvature.

6. Summary Checklist (AO1–AO3)

• Master the nine core differentiation rules (power, product, quotient, chain, exponential, logarithmic, trig, inverse‑trig, hyperbolic).
• Apply implicit and parametric differentiation confidently; obtain first and second derivatives.
• Use Leibniz’s rule for high‑order derivatives of products.
• Identify stationary points by solving \(f'(x)=0\); classify them with the second‑derivative test or the higher‑order test.
• Detect points of inflection via a sign change of \(f''\) or by confirming \(f'''(c)eq0\).
• Integrate these techniques into curve sketching, optimisation, related‑rates, and motion‑along‑a‑curve problems – the typical contexts of Cambridge A‑Level Paper 3.
• For Paper 1, be able to carry out the entire sketching process for pure algebraic functions (polynomials and rational functions) without invoking trigonometric or exponential forms.