Introduction to Differentiation
Gradient of a curve · First principles · Notation · §12.1
What differentiation tells us
The derivative \(\dfrac{dy}{dx}\) gives the gradient of the curve at any point. It is the limit of the gradient of the chord as the two points get infinitely close.
\(\dfrac{dy}{dx} = \lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\)
Notation: \(\dfrac{dy}{dx}\), \(f'(x)\), and \(y'\) all mean the same thing — the derivative with respect to x.
First principles example — differentiate \(f(x)=x^2\)
\(f(x+h) = (x+h)^2 = x^2+2xh+h^2\)
\(\dfrac{f(x+h)-f(x)}{h} = \dfrac{2xh+h^2}{h} = 2x+h\)
\(\lim_{h\to0}(2x+h) = 2x\)
\(\dfrac{f(x+h)-f(x)}{h} = \dfrac{2xh+h^2}{h} = 2x+h\)
\(\lim_{h\to0}(2x+h) = 2x\)
Result: \(\dfrac{d}{dx}(x^2) = 2x\)
Differentiation Rules
Power rule · Sum/difference · Constant multiple · §12.2
Power Rule
\(\dfrac{d}{dx}(x^n) = nx^{n-1}\)
Bring the power down and reduce it by 1. Works for any real n including fractions and negatives.
Constant
\(\dfrac{d}{dx}(c) = 0\)
The derivative of any constant is zero — it has no rate of change.
Sum/Difference
\(\dfrac{d}{dx}[f\pm g] = f'\pm g'\)
Differentiate each term separately. Works term by term across + and −.
Differentiate \(y = 5x^4 - 3x^2 + 7x - 2\)
Term by term
\(\dfrac{dy}{dx} = 20x^3 - 6x + 7\)
Note
Constant −2 gives zero, coefficient of x gives just 7
Differentiate \(y = \dfrac{3}{x^2} + 5\sqrt{x}\)
Rewrite
\(y = 3x^{-2} + 5x^{1/2}\)
Differentiate
\(\dfrac{dy}{dx} = -6x^{-3} + \dfrac{5}{2}x^{-1/2}\)
✓
\(= -\dfrac{6}{x^3} + \dfrac{5}{2\sqrt{x}}\)
Interactive — differentiation checker
Visualise \(y=ax^n\) and its derivative
—
Chain Rule
Differentiating composite functions · (f∘g)' = f'(g)·g' · §12.3
The chain rule
\(\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}\)
For composite functions \(y = f(g(x))\):
Let \(u = g(x)\) (the inner function), then differentiate both parts.
Shortcut: For \(y=(f(x))^n\): derivative = \(n(f(x))^{n-1} \cdot f'(x)\). Bring down n, reduce power by 1, multiply by derivative of inside.
Differentiate \(y = (3x^2+1)^5\)
Let u
\(u = 3x^2+1\), so \(y=u^5\)
dy/du
\(\dfrac{dy}{du} = 5u^4\)
du/dx
\(\dfrac{du}{dx} = 6x\)
✓
\(\dfrac{dy}{dx} = 5(3x^2+1)^4 \cdot 6x = 30x(3x^2+1)^4\)
Differentiate \(y = \sqrt{5x-3}\)
Rewrite
\(y = (5x-3)^{1/2}\)
Chain rule
\(\dfrac{dy}{dx} = \dfrac{1}{2}(5x-3)^{-1/2} \cdot 5\)
✓
\(= \dfrac{5}{2\sqrt{5x-3}}\)
Tangents & Normals
Equations of tangents · Normal (perpendicular) · Stationary points intro · §12.4
Tangent and normal at a point
Tangent gradient: \(m_T = f'(x_0)\) — substitute the x-value into the derivative
Normal gradient: \(m_N = -\dfrac{1}{m_T}\) — perpendicular to tangent
Both use: \(y - y_0 = m(x - x_0)\)
Finding stationary points: set \(\dfrac{dy}{dx}=0\) and solve. Substitute back into original equation to find y-coordinates.
Find tangent and normal to \(y=x^3-2x\) at \(x=2\)
y at x=2
\(y=8-4=4\) → point (2,4)
dy/dx
\(\dfrac{dy}{dx}=3x^2-2\)
Tangent m
At x=2: \(m_T=3(4)-2=10\)
Tangent
\(y-4=10(x-2)\Rightarrow y=10x-16\)
Normal m
\(m_N=-\tfrac{1}{10}\)
✓
Normal: \(y-4=-\tfrac{1}{10}(x-2)\Rightarrow y=-\tfrac{x}{10}+\tfrac{21}{5}\)
Stationary Points & Curve Sketching
Local max/min · Second derivative test · Nature of stationary points · §12.5
Finding and classifying stationary points
- Set \(\dfrac{dy}{dx}=0\) and solve to find x-coordinates of stationary points
- Substitute back into \(y\) to find y-coordinates
- Find \(\dfrac{d^2y}{dx^2}\) (second derivative)
- Substitute x-value: if >0 → minimum, if <0 → maximum, if =0 → inconclusive
\(\dfrac{d^2y}{dx^2} > 0 \Rightarrow \text{minimum}\quad\dfrac{d^2y}{dx^2} < 0 \Rightarrow \text{maximum}\)
Find stationary points of \(y=2x^3-9x^2+12x-4\)
dy/dx=0
\(6x^2-18x+12=0\Rightarrow x^2-3x+2=0\Rightarrow (x-1)(x-2)=0\)
x values
\(x=1\) and \(x=2\)
y values
\(y(1)=2-9+12-4=1\) and \(y(2)=16-36+24-4=0\)
d²y/dx²
\(12x-18\). At x=1: \(-6<0\) → MAX. At x=2: \(6>0\) → MIN.
✓
Local max (1,1); local min (2,0)
Practice Questions
3 Easy · 3 Medium · 3 Hard — Differentiation
Easy 1
[2] Differentiate \(y=4x^3-6x^2+2x-5\).
SOLUTION
Power rule
\(\dfrac{dy}{dx}=12x^2-12x+2\)
Easy 2
[2] Find the gradient of \(y=x^3-3x\) at the point where \(x=2\).
SOLUTION
Differentiate
\(\frac{dy}{dx}=3x^2-3\)
At x=2
\(3(4)-3=\boldsymbol{9}\)
Easy 3
[3] Find \(\dfrac{d}{dx}\big[(2x+1)^5\big]\).
SOLUTION — chain rule
Chain rule
\(5(2x+1)^4\times2=\boldsymbol{10(2x+1)^4}\)
Medium 1
[4] Differentiate \(y=x^2\sin x\).
SOLUTION — product rule
Let u, v
\(u=x^2,\;v=\sin x\Rightarrow u'=2x,\;v'=\cos x\)
Apply
\(\frac{dy}{dx}=uv'+vu'=x^2\cos x+2x\sin x\)
Medium 2
[4] Differentiate \(y=\dfrac{e^x}{x^2+1}\).
SOLUTION — quotient rule
u, v
\(u=e^x,\;v=x^2+1\Rightarrow u'=e^x,\;v'=2x\)
Apply
\(\dfrac{dy}{dx}=\dfrac{e^x(x^2+1)-e^x(2x)}{(x^2+1)^2}=\dfrac{e^x(x^2-2x+1)}{(x^2+1)^2}=\dfrac{e^x(x-1)^2}{(x^2+1)^2}\)
Medium 3
[4] Find the equation of the tangent to \(y=3x^2-x^3\) at the point \((2,4)\).
SOLUTION
Gradient
\(\frac{dy}{dx}=6x-3x^2\). At \(x=2\): \(12-12=0\)
Tangent
Horizontal tangent through \((2,4)\): \(\boldsymbol{y=4}\)
Hard 1
[5] Find the stationary points of \(y=x^3-6x^2+9x+2\) and determine their nature.
SOLUTION
dy/dx=0
\(3x^2-12x+9=0\Rightarrow x^2-4x+3=0\Rightarrow(x-1)(x-3)=0\)
Points
\(x=1: y=6\) \(x=3: y=2\)
Nature (d²y/dx²)
\(\frac{d^2y}{dx^2}=6x-12\)
\(x=1: -6<0\Rightarrow\) Local max (1,6)
\(x=3: 6>0\Rightarrow\) Local min (3,2)
\(x=1: -6<0\Rightarrow\) Local max (1,6)
\(x=3: 6>0\Rightarrow\) Local min (3,2)
Hard 2
[5] Differentiate \(y=\ln(\cos x)\) and \(y=e^{x^2}\). Hence find the equation of the normal to \(y=e^{x^2}\) at \(x=1\).
SOLUTION
\(\ln(\cos x)\)
Chain rule: \(\dfrac{1}{\cos x}\times(-\sin x)=-\tan x\)
\(e^{x^2}\)
Chain rule: \(e^{x^2}\times2x=2xe^{x^2}\)
Normal at x=1
Gradient of curve \(=2e\). Normal gradient \(=-\dfrac{1}{2e}\)
Point: \((1,e)\). Normal: \(y-e=-\dfrac{1}{2e}(x-1)\)
Point: \((1,e)\). Normal: \(y-e=-\dfrac{1}{2e}(x-1)\)
Hard 3
[6] A curve is defined implicitly by \(x^2+3xy+y^2=11\). Find \(\dfrac{dy}{dx}\) in terms of \(x\) and \(y\). Hence find the coordinates of the points where the tangent is parallel to the x-axis.
SOLUTION — implicit differentiation
Differentiate
\(2x+3y+3x\frac{dy}{dx}+2y\frac{dy}{dx}=0\)
Rearrange
\(\frac{dy}{dx}(3x+2y)=-(2x+3y)\Rightarrow\dfrac{dy}{dx}=\dfrac{-(2x+3y)}{3x+2y}\)
Horizontal tangent
\(\frac{dy}{dx}=0\Rightarrow 2x+3y=0\Rightarrow y=-\frac{2x}{3}\)
Sub: \(x^2+3x(-\frac{2x}{3})+\frac{4x^2}{9}=11\Rightarrow x^2(1-2+\frac49)=11\Rightarrow x^2\cdot\frac{-5}{9}=11\)
No real solution — tangent is never horizontal on this ellipse.
Sub: \(x^2+3x(-\frac{2x}{3})+\frac{4x^2}{9}=11\Rightarrow x^2(1-2+\frac49)=11\Rightarrow x^2\cdot\frac{-5}{9}=11\)
No real solution — tangent is never horizontal on this ellipse.
Past Year Paper Questions
Cambridge 0606 — Differentiation
0606
[11]
The curve \(C\) has equation \(y = (x-2)^2(x+1)\).
(a) Find \(\frac{dy}{dx}\) and the coordinates of the stationary points. [5]
(b) Determine the nature of each stationary point. [3]
(c) Sketch curve \(C\), marking intercepts and stationary points. [3]
(a) Find \(\frac{dy}{dx}\) and the coordinates of the stationary points. [5]
(b) Determine the nature of each stationary point. [3]
(c) Sketch curve \(C\), marking intercepts and stationary points. [3]
FULL SOLUTION
(a) Expand
\(y=x^3-3x^2+4\). \(\frac{dy}{dx}=3x^2-6x=3x(x-2)=0\Rightarrow x=0\) or \(x=2\)
Points: \((0,4)\) and \((2,0)\)
Points: \((0,4)\) and \((2,0)\)
(b) Nature
\(\frac{d^2y}{dx^2}=6x-6\)
\(x=0:-6<0\Rightarrow\) local max \((0,4)\)
\(x=2:6>0\Rightarrow\) local min \((2,0)\)
\(x=0:-6<0\Rightarrow\) local max \((0,4)\)
\(x=2:6>0\Rightarrow\) local min \((2,0)\)
(c)
x-intercepts: \(x=2\) (touches), \(x=-1\) (crosses). y-intercept: 4. Max at \((0,4)\), min at \((2,0)\).
0606 Style
Implicit & Parametric Differentiation
[8]
A curve is defined parametrically by \(x = t^2 + 1\), \(y = 2t^3 - 3t\).
(a) Find \(\dfrac{dy}{dx}\) in terms of \(t\). [3]
(b) Find the gradient of the curve at \(t = 2\). [2]
(c) Find the coordinates and equation of the tangent at \(t = 1\). [3]
(a) Find \(\dfrac{dy}{dx}\) in terms of \(t\). [3]
(b) Find the gradient of the curve at \(t = 2\). [2]
(c) Find the coordinates and equation of the tangent at \(t = 1\). [3]
FULL WORKED SOLUTION
(a)
\(\frac{dx}{dt}=2t\), \(\frac{dy}{dt}=6t^2-3\).
\(\frac{dy}{dx}=\frac{6t^2-3}{2t}=\boldsymbol{3t-\frac{3}{2t}}\)
\(\frac{dy}{dx}=\frac{6t^2-3}{2t}=\boldsymbol{3t-\frac{3}{2t}}\)
(b)
At \(t=2\): \(\frac{dy}{dx}=\frac{6(4)-3}{4}=\frac{21}{4}=\boldsymbol{5.25}\)
(c)
At \(t=1\): \(x=2, y=-1\). Gradient \(=\frac{3}{2}\).
Tangent: \(y+1=\frac{3}{2}(x-2)\Rightarrow\boldsymbol{2y=3x-8}\)
Tangent: \(y+1=\frac{3}{2}(x-2)\Rightarrow\boldsymbol{2y=3x-8}\)
0606 Style
Second Derivative & Concavity
[7]
\(f(x) = \ln(3x^2+1)\).
(a) Find \(f'(x)\) and \(f''(x)\). [4]
(b) Find the x-values where the graph of \(y=f(x)\) changes concavity (points of inflection). [3]
(a) Find \(f'(x)\) and \(f''(x)\). [4]
(b) Find the x-values where the graph of \(y=f(x)\) changes concavity (points of inflection). [3]
FULL WORKED SOLUTION
(a)
\(f'(x)=\dfrac{6x}{3x^2+1}\)
Quotient rule: \(f''(x)=\dfrac{6(3x^2+1)-6x(6x)}{(3x^2+1)^2}=\dfrac{6-18x^2}{(3x^2+1)^2}\)
Quotient rule: \(f''(x)=\dfrac{6(3x^2+1)-6x(6x)}{(3x^2+1)^2}=\dfrac{6-18x^2}{(3x^2+1)^2}\)
(b)
\(f''=0\Rightarrow 6-18x^2=0\Rightarrow x^2=\frac{1}{3}\Rightarrow\boldsymbol{x=\pm\frac{1}{\sqrt3}}\)
Sign change confirmed → points of inflection at \(x=\pm\frac{1}{\sqrt3}\)
Sign change confirmed → points of inflection at \(x=\pm\frac{1}{\sqrt3}\)