Limits and Continuity Intuitive Notion of a Limit The limit of a function f ( x ) f(x) f ( x ) x x x a a a f ( x ) f(x) f ( x ) f ( a ) f(a) f ( a )
lim x → a f ( x ) = L \lim_{x \to a} f(x) = L x → a lim f ( x ) = L This means that as x x x a a a f ( x ) f(x) f ( x ) L L L
Left-Hand and Right-Hand Limits A two-sided limit exists if and only if both one-sided limits exist and are equal:
lim x → a f ( x ) = L ⟺ lim x → a − f ( x ) = lim x → a + f ( x ) = L \lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L x → a lim f ( x ) = L ⟺ x → a − lim f ( x ) = x → a + lim f ( x ) = L Example
Find lim x → 0 ∣ x ∣ x \displaystyle\lim_{x \to 0} \frac{|x|}{x} x → 0 lim x ∣ x ∣
lim x → 0 − ∣ x ∣ x = lim x → 0 − − x x = − 1 \lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1 x → 0 − lim x ∣ x ∣ = x → 0 − lim x − x = − 1 lim x → 0 + ∣ x ∣ x = lim x → 0 + x x = 1 \lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1 x → 0 + lim x ∣ x ∣ = x → 0 + lim x x = 1 Since the one-sided limits are not equal, the limit does not exist.
Limit Value lim x → 0 sin x x \displaystyle\lim_{x \to 0} \frac{\sin x}{x} x → 0 lim x sin x 1 1 1 lim x → 0 1 − cos x x \displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x} x → 0 lim x 1 − cos x 0 0 0 lim x → ∞ 1 x \displaystyle\lim_{x \to \infty} \frac{1}{x} x → ∞ lim x 1 0 0 0 lim x → ∞ ( 1 + 1 x ) x \displaystyle\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x x → ∞ lim ( 1 + x 1 ) x e e e lim x → 0 e x − 1 x \displaystyle\lim_{x \to 0} \frac{e^x - 1}{x} x → 0 lim x e x − 1 1 1 1 lim x → 0 ln ( 1 + x ) x \displaystyle\lim_{x \to 0} \frac{\ln(1+x)}{x} x → 0 lim x ln ( 1 + x ) 1 1 1
The Squeeze Theorem If g ( x ) ≤ f ( x ) ≤ h ( x ) g(x) \le f(x) \le h(x) g ( x ) ≤ f ( x ) ≤ h ( x ) x x x a a a a a a
lim x → a g ( x ) = lim x → a h ( x ) = L \lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L x → a lim g ( x ) = x → a lim h ( x ) = L Then lim x → a f ( x ) = L \displaystyle\lim_{x \to a} f(x) = L x → a lim f ( x ) = L
Example
Show that lim x → 0 x 2 sin ( 1 x ) = 0 \displaystyle\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0 x → 0 lim x 2 sin ( x 1 ) = 0
Since − 1 ≤ sin ( 1 x ) ≤ 1 -1 \le \sin\!\left(\frac{1}{x}\right) \le 1 − 1 ≤ sin ( x 1 ) ≤ 1 − x 2 ≤ x 2 sin ( 1 x ) ≤ x 2 -x^2 \le x^2 \sin\!\left(\frac{1}{x}\right) \le x^2 − x 2 ≤ x 2 sin ( x 1 ) ≤ x 2
Both lim x → 0 ( − x 2 ) = 0 \displaystyle\lim_{x \to 0}(-x^2) = 0 x → 0 lim ( − x 2 ) = 0 lim x → 0 x 2 = 0 \displaystyle\lim_{x \to 0} x^2 = 0 x → 0 lim x 2 = 0
By the squeeze theorem, lim x → 0 x 2 sin ( 1 x ) = 0 \displaystyle\lim_{x \to 0} x^2 \sin\!\left(\frac{1}{x}\right) = 0 x → 0 lim x 2 sin ( x 1 ) = 0
Continuity A function f f f continuous at x = a x = a x = a
f ( a ) f(a) f ( a ) lim x → a f ( x ) \displaystyle\lim_{x \to a} f(x) x → a lim f ( x ) lim x → a f ( x ) = f ( a ) \displaystyle\lim_{x \to a} f(x) = f(a) x → a lim f ( x ) = f ( a )
When asked whether a piecewise function is continuous at a boundary point, always check that the Left-hand limit, right-hand limit, and function value all agree.
The Derivative Definition from First Principles The derivative of f f f x = a x = a x = a
F ′ ( a ) = lim h → 0 f ( a + h ) − f ( a ) h F'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} F ′ ( a ) = h → 0 lim h f ( a + h ) − f ( a ) This limit, when it exists, gives the instantaneous rate of change of f f f x = a x = a x = a y = f ( x ) y = f(x) y = f ( x )
F ′ ( x ) = lim Δ x → 0 f ( x + Δ x ) − f ( x ) Δ x F'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} F ′ ( x ) = Δ x → 0 lim Δ x f ( x + Δ x ) − f ( x ) Differentiability Implies Continuity If f f f x = a x = a x = a f f f x = a x = a x = a f ( x ) = ∣ x ∣ f(x) = |x| f ( x ) = ∣ x ∣ x = 0 x = 0 x = 0
Example
Differentiate f ( x ) = x 2 f(x) = x^2 f ( x ) = x 2
\begin`\{aligned}` F'(x) &= \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} \\[6pt] &= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} \\[6pt] &= \lim_{h \to 0} \frac{2xh + h^2}{h} \\[6pt] &= \lim_{h \to 0}(2x + h) \\[6pt] &= 2x \end`\{aligned}` Notation for Derivatives Several notations are equivalent:
f ′ ( x ) f'(x) f ′ ( x ) d y d x \dfrac{dy}{dx} d x d y d d x [ f ( x ) ] \dfrac{d}{dx}[f(x)] d x d [ f ( x )] y ˙ \dot{y} y ˙ Differentiation Rules The Power Rule For n ∈ R n \in \mathbb{R} n ∈ R
\frac{d}`\{dx}`[x^n] = nx^{n-1} Constant Multiple Rule \frac{d}`\{dx}`[cf(x)] = c \cdot f'(x) Sum and Difference Rule \frac{d}`\{dx}`[f(x) \pm g(x)] = f'(x) \pm g'(x) The Product Rule If u = f ( x ) u = f(x) u = f ( x ) v = g ( x ) v = g(x) v = g ( x )
\frac{d}`\{dx}`[uv] = u\frac`\{dv}``\{dx}` + v\frac`\{du}``\{dx}` Example
Differentiate f ( x ) = x 3 sin x f(x) = x^3 \sin x f ( x ) = x 3 sin x
F ′ ( x ) = 3 x 2 sin x + x 3 cos x F'(x) = 3x^2 \sin x + x^3 \cos x F ′ ( x ) = 3 x 2 sin x + x 3 cos x The Quotient Rule \frac{d}`\{dx}`\left[\frac{u}{v}\right] = \frac{v\frac`\{du}``\{dx}` - u\frac`\{dv}``\{dx}`}{v^2} Example
Differentiate f ( x ) = x 2 + 1 x − 3 \displaystyle f(x) = \frac{x^2 + 1}{x - 3} f ( x ) = x − 3 x 2 + 1
Let u = x 2 + 1 u = x^2 + 1 u = x 2 + 1 v = x − 3 v = x - 3 v = x − 3 u ′ = 2 x u' = 2x u ′ = 2 x v ′ = 1 v' = 1 v ′ = 1
\begin`\{aligned}` F'(x) &= \frac{(x-3)(2x) - (x^2+1)(1)}{(x-3)^2} \\[6pt] &= \frac{2x^2 - 6x - x^2 - 1}{(x-3)^2} \\[6pt] &= \frac{x^2 - 6x - 1}{(x-3)^2} \end`\{aligned}` The Chain Rule If y = f ( g ( x ) ) y = f(g(x)) y = f ( g ( x ))
\frac`\{dy}``\{dx}` = f'(g(x)) \cdot g'(x) Or in Leibniz notation, if y = f ( u ) y = f(u) y = f ( u ) u = g ( x ) u = g(x) u = g ( x )
\frac`\{dy}``\{dx}` = \frac`\{dy}``\{du}` \cdot \frac`\{du}``\{dx}` Example
Differentiate f ( x ) = ( 3 x 2 + 1 ) 5 f(x) = (3x^2 + 1)^5 f ( x ) = ( 3 x 2 + 1 ) 5
Let u = 3 x 2 + 1 u = 3x^2 + 1 u = 3 x 2 + 1 y = u 5 y = u^5 y = u 5
\frac`\{dy}``\{dx}` = 5u^4 \cdot 6x = 30x(3x^2+1)^4 Example
Differentiate f ( x ) = sin ( 2 x 2 + 1 ) \displaystyle f(x) = \sin(2x^2 + 1) f ( x ) = sin ( 2 x 2 + 1 )
Let u = 2 x 2 + 1 u = 2x^2 + 1 u = 2 x 2 + 1 y = sin u y = \sin u y = sin u
\frac`\{dy}``\{dx}` = \cos u \cdot 4x = 4x\cos(2x^2+1) Derivatives of Standard Functions f ( x ) f(x) f ( x ) f ′ ( x ) f'(x) f ′ ( x ) x n x^n x n n x n − 1 nx^{n-1} n x n − 1 e x e^x e x e x e^x e x a x a^x a x a x ln a a^x \ln a a x ln a ln x \ln x ln x 1 x \dfrac{1}{x} x 1 log a x \log_a x log a x 1 x ln a \dfrac{1}{x \ln a} x ln a 1 sin x \sin x sin x cos x \cos x cos x cos x \cos x cos x − sin x -\sin x − sin x tan x \tan x tan x sec 2 x \sec^2 x sec 2 x csc x \csc x csc x − csc x cot x -\csc x \cot x − csc x cot x sec x \sec x sec x sec x tan x \sec x \tan x sec x tan x cot x \cot x cot x − csc 2 x -\csc^2 x − csc 2 x arcsin x \arcsin x arcsin x 1 1 − x 2 \dfrac{1}{\sqrt{1-x^2}} 1 − x 2 1 arccos x \arccos x arccos x − 1 1 − x 2 \dfrac{-1}{\sqrt{1-x^2}} 1 − x 2 − 1 arctan x \arctan x arctan x 1 1 + x 2 \dfrac{1}{1+x^2} 1 + x 2 1
The IB formula booklet provides the derivatives of \sin x$$\cos x$$\tan x$$e^x And ln x \ln x ln x
Implicit Differentiation When a function is not given explicitly as y = f ( x ) y = f(x) y = f ( x ) x x x y y y x x x d y d x \dfrac{dy}{dx} d x d y
Key Idea When differentiating a term involving y y y x x x
\frac{d}`\{dx}`[y^n] = ny^{n-1}\frac`\{dy}``\{dx}` Example
Find d y d x \dfrac{dy}{dx} d x d y x 2 + y 2 = 25 x^2 + y^2 = 25 x 2 + y 2 = 25
Differentiate both sides with respect to x x x
2x + 2y\frac`\{dy}``\{dx}` = 0 \frac`\{dy}``\{dx}` = -\frac{x}{y} At the point ( 3 , 4 ) (3, 4) ( 3 , 4 ) d y d x = − 3 4 \dfrac{dy}{dx} = -\dfrac{3}{4} d x d y = − 4 3
Example
Find d y d x \dfrac{dy}{dx} d x d y x 3 + y 3 = 6 x y x^3 + y^3 = 6xy x 3 + y 3 = 6 x y
3x^2 + 3y^2\frac`\{dy}``\{dx}` = 6y + 6x\frac`\{dy}``\{dx}` 3y^2\frac`\{dy}``\{dx}` - 6x\frac`\{dy}``\{dx}` = 6y - 3x^2 \frac`\{dy}``\{dx}`(3y^2 - 6x) = 6y - 3x^2 \frac`\{dy}``\{dx}` = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x} Second Derivatives Implicitly To find d 2 y d x 2 \dfrac{d^2y}{dx^2} d x 2 d 2 y d y d x \dfrac{dy}{dx} d x d y d y d x \dfrac{dy}{dx} d x d y x x x y y y
Example
Find d 2 y d x 2 \dfrac{d^2y}{dx^2} d x 2 d 2 y x 2 + y 2 = 25 x^2 + y^2 = 25 x 2 + y 2 = 25
We have d y d x = − x y \dfrac{dy}{dx} = -\dfrac{x}{y} d x d y = − y x
Differentiate with respect to x x x
\frac{d^2y}{dx^2} = -\frac{y \cdot 1 - x \cdot \frac`\{dy}``\{dx}`}{y^2} Substitute d y d x = − x y \dfrac{dy}{dx} = -\dfrac{x}{y} d x d y = − y x
d 2 y d x 2 = − y − x ( − x y ) y 2 = − y + x 2 y y 2 = − y 2 + x 2 y 3 = − 25 y 3 \frac{d^2y}{dx^2} = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{y^2 + x^2}{y^3} = -\frac{25}{y^3} d x 2 d 2 y = − y 2 y − x ( − y x ) = − y 2 y + y x 2 = − y 3 y 2 + x 2 = − y 3 25 When two or more quantities are related by an equation, their rates of change are also related.
Strategy Identify the quantities that are changing and the rate(s) given. Write an equation relating the quantities. Differentiate both sides with respect to time (t t t Substitute known values and solve for the unknown rate. Example
A spherical balloon is being inflated at a rate of 10 c m 3 / s 10\mathrm{ cm}^3\mathrm{/s} 10 cm 3 /s 5 c m 5\mathrm{ cm} 5 cm
Volume of a sphere: V = 4 3 π r 3 V = \dfrac{4}{3}\pi r^3 V = 3 4 π r 3
Differentiate with respect to t t t
\frac`\{dV}``\{dt}` = 4\pi r^2 \frac`\{dr}``\{dt}` Substitute d V d t = 10 \dfrac{dV}{dt} = 10 d t d V = 10 r = 5 r = 5 r = 5
10 = 4\pi(25)\frac`\{dr}``\{dt}` \frac`\{dr}``\{dt}` = \frac{10}{100\pi} = \frac{1}{10\pi} \approx 0.0318 \mathrm{ cm/s} Example
A ladder 10 m 10\mathrm{ m} 10 m 1 m / s 1\mathrm{ m/s} 1 m/s 6 m 6\mathrm{ m} 6 m
By Pythagoras: x 2 + y 2 = 100 x^2 + y^2 = 100 x 2 + y 2 = 100
Differentiate: 2 x d x d t + 2 y d y d t = 0 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0 2 x d t d x + 2 y d t d y = 0
When x = 6 x = 6 x = 6 y = 100 − 36 = 8 y = \sqrt{100-36} = 8 y = 100 − 36 = 8
2(6)(1) + 2(8)\frac`\{dy}``\{dt}` = 0 \frac`\{dy}``\{dt}` = -\frac{12}{16} = -0.75 \mathrm{ m/s} The negative sign means the top is sliding down.
Second Derivatives and Higher Derivatives Notation F''(x) = \frac{d^2y}{dx^2} = \frac{d}`\{dx}`\left(\frac`\{dy}``\{dx}`\right) F'''(x) = \frac{d^3y}{dx^3} = \frac{d}`\{dx}`\left(\frac{d^2y}{dx^2}\right) Example
Find f ′ ′ ( x ) f''(x) f ′′ ( x ) f ( x ) = x 4 − 3 x 3 + 2 x − 7 f(x) = x^4 - 3x^3 + 2x - 7 f ( x ) = x 4 − 3 x 3 + 2 x − 7
F ′ ( x ) = 4 x 3 − 9 x 2 + 2 F'(x) = 4x^3 - 9x^2 + 2 F ′ ( x ) = 4 x 3 − 9 x 2 + 2 F ′ ′ ( x ) = 12 x 2 − 18 x F''(x) = 12x^2 - 18x F ′′ ( x ) = 12 x 2 − 18 x F ′ ′ ′ ( x ) = 24 x − 18 F'''(x) = 24x - 18 F ′′′ ( x ) = 24 x − 18 Applications of Differentiation Tangents and Normals The tangent to y = f ( x ) y = f(x) y = f ( x ) x = a x = a x = a f ′ ( a ) f'(a) f ′ ( a )
Y − f ( a ) = f ′ ( a ) ( x − a ) Y - f(a) = f'(a)(x - a) Y − f ( a ) = f ′ ( a ) ( x − a ) The normal is perpendicular to the tangent, so its gradient is − 1 f ′ ( a ) -\dfrac{1}{f'(a)} − f ′ ( a ) 1 f ′ ( a ) ≠ 0 f'(a) \neq 0 f ′ ( a ) = 0
Y − f ( a ) = − 1 f ′ ( a ) ( x − a ) Y - f(a) = -\frac{1}{f'(a)}(x - a) Y − f ( a ) = − f ′ ( a ) 1 ( x − a ) Example
Find the equation of the tangent and normal to y = x 3 − 3 x + 2 y = x^3 - 3x + 2 y = x 3 − 3 x + 2 x = 1 x = 1 x = 1
At x = 1 x = 1 x = 1 y = 1 − 3 + 2 = 0 y = 1 - 3 + 2 = 0 y = 1 − 3 + 2 = 0 y ′ = 3 x 2 − 3 = 0 y' = 3x^2 - 3 = 0 y ′ = 3 x 2 − 3 = 0
Since y ′ = 0 y' = 0 y ′ = 0 y = 0 y = 0 y = 0
The normal is vertical: x = 1 x = 1 x = 1
Increasing and Decreasing Functions f f f increasing on an interval if f ′ ( x ) > 0 f'(x) \gt 0 f ′ ( x ) > 0 x x x f f f decreasing on an interval if f ′ ( x ) < 0 f'(x) \lt 0 f ′ ( x ) < 0 x x x f f f stationary at x = a x = a x = a f ′ ( a ) = 0 f'(a) = 0 f ′ ( a ) = 0 Stationary Points A stationary point occurs where f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0
Type First Derivative Test Second Derivative Test Local maximum f ′ f' f ′ + + + − - − f ′ ′ ( x ) < 0 f''(x) \lt 0 f ′′ ( x ) < 0 Local minimum f ′ f' f ′ − - − + + + f ′ ′ ( x ) > 0 f''(x) \gt 0 f ′′ ( x ) > 0 Point of inflection f ′ f' f ′ f ′ ′ ( x ) = 0 f''(x) = 0 f ′′ ( x ) = 0
Example
Find and classify the stationary points of f ( x ) = x 3 − 6 x 2 + 9 x + 1 f(x) = x^3 - 6x^2 + 9x + 1 f ( x ) = x 3 − 6 x 2 + 9 x + 1
F ′ ( x ) = 3 x 2 − 12 x + 9 = 3 ( x 2 − 4 x + 3 ) = 3 ( x − 1 ) ( x − 3 ) F'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3) F ′ ( x ) = 3 x 2 − 12 x + 9 = 3 ( x 2 − 4 x + 3 ) = 3 ( x − 1 ) ( x − 3 ) Setting f ′ ( x ) = 0 f'(x) = 0 f ′ ( x ) = 0 x = 1 x = 1 x = 1 x = 3 x = 3 x = 3
F ′ ′ ( x ) = 6 x − 12 F''(x) = 6x - 12 F ′′ ( x ) = 6 x − 12 At x = 1 x = 1 x = 1 f ′ ′ ( 1 ) = − 6 < 0 f''(1) = -6 \lt 0 f ′′ ( 1 ) = − 6 < 0 f ( 1 ) = 1 − 6 + 9 + 1 = 5 f(1) = 1 - 6 + 9 + 1 = 5 f ( 1 ) = 1 − 6 + 9 + 1 = 5 ( 1 , 5 ) (1, 5) ( 1 , 5 )
At x = 3 x = 3 x = 3 f ′ ′ ( 3 ) = 6 > 0 f''(3) = 6 \gt 0 f ′′ ( 3 ) = 6 > 0 f ( 3 ) = 27 − 54 + 27 + 1 = 1 f(3) = 27 - 54 + 27 + 1 = 1 f ( 3 ) = 27 − 54 + 27 + 1 = 1 ( 3 , 1 ) (3, 1) ( 3 , 1 )
Concavity and Points of Inflection Stationary Points and Concavity
Use the sliders to adjust coefficients of a polynomial and observe how f ′ ( x ) f'(x) f ′ ( x ) f ′ ′ ( x ) f''(x) f ′′ ( x )
f f f concave up on an interval if f ′ ′ ( x ) > 0 f''(x) \gt 0 f ′′ ( x ) > 0 f f f concave down on an interval if f ′ ′ ( x ) < 0 f''(x) \lt 0 f ′′ ( x ) < 0 A point of inflection occurs where the concavity changes, i.e., f ′ ′ ( x ) f''(x) f ′′ ( x ) Example
Find the points of inflection of f ( x ) = x 4 − 4 x 3 f(x) = x^4 - 4x^3 f ( x ) = x 4 − 4 x 3
F ′ ( x ) = 4 x 3 − 12 x 2 F'(x) = 4x^3 - 12x^2 F ′ ( x ) = 4 x 3 − 12 x 2 F ′ ′ ( x ) = 12 x 2 − 24 x = 12 x ( x − 2 ) F''(x) = 12x^2 - 24x = 12x(x - 2) F ′′ ( x ) = 12 x 2 − 24 x = 12 x ( x − 2 ) f ′ ′ ( x ) = 0 f''(x) = 0 f ′′ ( x ) = 0 x = 0 x = 0 x = 0 x = 2 x = 2 x = 2
Interval Sign of f ′ ′ f'' f ′′ Concavity x < 0 x \lt 0 x < 0 + + + Concave up 0 < x < 2 0 \lt x \lt 2 0 < x < 2 − - − Concave down x > 2 x \gt 2 x > 2 + + + Concave up
Concavity changes at both x = 0 x = 0 x = 0 x = 2 x = 2 x = 2
At x = 0 x = 0 x = 0 f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 ( 0 , 0 ) (0, 0) ( 0 , 0 )
At x = 2 x = 2 x = 2 f ( 2 ) = 16 − 32 = − 16 f(2) = 16 - 32 = -16 f ( 2 ) = 16 − 32 = − 16 ( 2 , − 16 ) (2, -16) ( 2 , − 16 )
Optimization Optimization problems involve finding the maximum or minimum value of a quantity subject to Constraints.
Strategy Define variables and identify the quantity to be optimised. Write an expression for the quantity in terms of a single variable. Differentiate and set the derivative equal to zero. Verify that the critical point gives a maximum or minimum. Answer the question in context. Example
A piece of wire 100 c m 100\mathrm{ cm} 100 cm
Let the dimensions be x x x y y y 2 x + 2 y = 100 2x + 2y = 100 2 x + 2 y = 100 y = 50 − x y = 50 - x y = 50 − x
Area: A = x y = x ( 50 − x ) = 50 x − x 2 A = xy = x(50 - x) = 50x - x^2 A = x y = x ( 50 − x ) = 50 x − x 2
\frac`\{dA}``\{dx}` = 50 - 2x = 0 \implies x = 25 d 2 A d x 2 = − 2 < 0 ⟹ m a x i m u m \frac{d^2A}{dx^2} = -2 \lt 0 \implies \mathrm{maximum} d x 2 d 2 A = − 2 < 0 ⟹ maximum So x = 25\mathrm{ cm}$$y = 25\mathrm{ cm} . The rectangle is a square with area 625 c m 2 625\mathrm{ cm}^2 625 cm 2
Example
An open-top cylindrical can is to hold 500 c m 3 500\mathrm{ cm}^3 500 cm 3
Volume: V = π r 2 h = 500 V = \pi r^2 h = 500 V = π r 2 h = 500 h = 500 π r 2 h = \dfrac{500}{\pi r^2} h = π r 2 500
Surface area (no top): A = π r 2 + 2 π r h = π r 2 + 1000 r A = \pi r^2 + 2\pi r h = \pi r^2 + \dfrac{1000}{r} A = π r 2 + 2 π r h = π r 2 + r 1000
\frac`\{dA}``\{dr}` = 2\pi r - \frac{1000}{r^2} = 0 2 π r = 1000 r 2 ⟹ 2 π r 3 = 1000 ⟹ r 3 = 500 π 2\pi r = \frac{1000}{r^2} \implies 2\pi r^3 = 1000 \implies r^3 = \frac{500}{\pi} 2 π r = r 2 1000 ⟹ 2 π r 3 = 1000 ⟹ r 3 = π 500 R = ( 500 π ) 1 / 3 ≈ 5.42 c m R = \left(\frac{500}{\pi}\right)^{1/3} \approx 5.42 \mathrm{ cm} R = ( π 500 ) 1/3 ≈ 5.42 cm H = 500 π ⋅ ( 500 π ) 2 / 3 = ( 500 π ) 1 / 3 ≈ 5.42 c m H = \frac{500}{\pi \cdot \left(\frac{500}{\pi}\right)^{2/3}} = \left(\frac{500}{\pi}\right)^{1/3} \approx 5.42 \mathrm{ cm} H = π ⋅ ( π 500 ) 2/3 500 = ( π 500 ) 1/3 ≈ 5.42 cm The optimal open-top can has h = r h = r h = r
L’Hopital’s Rule If lim x → a f ( x ) g ( x ) \displaystyle\lim_{x \to a}\frac{f(x)}{g(x)} x → a lim g ( x ) f ( x ) 0 0 \dfrac{0}{0} 0 0 ± ∞ ± ∞ \dfrac{\pm\infty}{\pm\infty} ± ∞ ± ∞
lim x → a f ( x ) g ( x ) = lim x → a f ′ ( x ) g ′ ( x ) \lim_{x \to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{f'(x)}{g'(x)} x → a lim g ( x ) f ( x ) = x → a lim g ′ ( x ) f ′ ( x ) Provided the limit on the right exists.
Example
Evaluate lim x → 0 e x − 1 − x x 2 \displaystyle\lim_{x \to 0}\frac{e^x - 1 - x}{x^2} x → 0 lim x 2 e x − 1 − x
Direct substitution gives 0 0 \dfrac{0}{0} 0 0
lim x → 0 e x − 1 2 x \lim_{x \to 0}\frac{e^x - 1}{2x} x → 0 lim 2 x e x − 1 Still 0 0 \dfrac{0}{0} 0 0
lim x → 0 e x 2 = 1 2 \lim_{x \to 0}\frac{e^x}{2} = \frac{1}{2} x → 0 lim 2 e x = 2 1 IB Exam-Style Questions Question 1 (Paper 1 style) Let f ( x ) = 2 x 3 − 3 x 2 − 36 x + 5 f(x) = 2x^3 - 3x^2 - 36x + 5 f ( x ) = 2 x 3 − 3 x 2 − 36 x + 5
(a) Find f ′ ( x ) f'(x) f ′ ( x )
F ′ ( x ) = 6 x 2 − 6 x − 36 F'(x) = 6x^2 - 6x - 36 F ′ ( x ) = 6 x 2 − 6 x − 36 (b) Find the x x x
6 x 2 − 6 x − 36 = 0 ⟹ x 2 − x − 6 = 0 ⟹ ( x − 3 ) ( x + 2 ) = 0 6x^2 - 6x - 36 = 0 \implies x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0 6 x 2 − 6 x − 36 = 0 ⟹ x 2 − x − 6 = 0 ⟹ ( x − 3 ) ( x + 2 ) = 0 x = 3 x = 3 x = 3 x = − 2 x = -2 x = − 2
(c) Determine the nature of each stationary point.
F ′ ′ ( x ) = 12 x − 6 F''(x) = 12x - 6 F ′′ ( x ) = 12 x − 6 At x = − 2 x = -2 x = − 2 f ′ ′ ( − 2 ) = − 30 < 0 f''(-2) = -30 \lt 0 f ′′ ( − 2 ) = − 30 < 0
At x = 3 x = 3 x = 3 f ′ ′ ( 3 ) = 30 > 0 f''(3) = 30 \gt 0 f ′′ ( 3 ) = 30 > 0
(d) Find the equation of the tangent to the curve at x = 0 x = 0 x = 0
f ( 0 ) = 5 f(0) = 5 f ( 0 ) = 5 f ′ ( 0 ) = − 36 f'(0) = -36 f ′ ( 0 ) = − 36
Y − 5 = − 36 ( x − 0 ) ⟹ y = − 36 x + 5 Y - 5 = -36(x - 0) \implies y = -36x + 5 Y − 5 = − 36 ( x − 0 ) ⟹ y = − 36 x + 5 Question 2 (Paper 2 style) A rectangular garden is to be constructed adjacent to a house, with the house forming one side of The rectangle. Fencing is required for the other three sides. 60 m 60\mathrm{ m} 60 m
(a) If the side parallel to the house has length x m x\mathrm{ m} x m
A = x ( 30 − x 2 ) = 30 x − x 2 2 A = x\left(30 - \frac{x}{2}\right) = 30x - \frac{x^2}{2} A = x ( 30 − 2 x ) = 30 x − 2 x 2 The two perpendicular sides have total length 60 − x 60 - x 60 − x 60 − x 2 = 30 − x 2 \dfrac{60-x}{2} = 30 - \dfrac{x}{2} 2 60 − x = 30 − 2 x
A = x ( 30 − x 2 ) A = x\left(30 - \frac{x}{2}\right) A = x ( 30 − 2 x ) (b) Find the value of x x x
\frac`\{dA}``\{dx}` = 30 - x = 0 \implies x = 30 d 2 A d x 2 = − 1 < 0 ⟹ m a x i m u m \frac{d^2A}{dx^2} = -1 \lt 0 \implies \mathrm{maximum} d x 2 d 2 A = − 1 < 0 ⟹ maximum Maximum area = 30 × 15 = 450 m 2 = 30 \times 15 = 450\mathrm{ m}^2 = 30 × 15 = 450 m 2
Question 3 (Paper 1 style) Given x 2 + x y + y 2 = 7 x^2 + xy + y^2 = 7 x 2 + x y + y 2 = 7
(a) Find d y d x \dfrac{dy}{dx} d x d y x x x y y y
2x + y + x\frac`\{dy}``\{dx}` + 2y\frac`\{dy}``\{dx}` = 0 (x + 2y)\frac`\{dy}``\{dx}` = -(2x + y) \frac`\{dy}``\{dx}` = -\frac{2x+y}{x+2y} (b) Verify that the point ( 1 , 2 ) (1, 2) ( 1 , 2 )
1 + 2 + 4 = 7 1 + 2 + 4 = 7 1 + 2 + 4 = 7
\frac`\{dy}``\{dx}`\bigg|_{(1,2)} = -\frac{2+2}{1+4} = -\frac{4}{5} Question 4 (Paper 2 style) The curve C C C y = x e − x y = x e^{-x} y = x e − x
(a) Find d y d x \dfrac{dy}{dx} d x d y
Using the product rule with u = x u = x u = x v = e − x v = e^{-x} v = e − x
\frac`\{dy}``\{dx}` = 1 \cdot e^{-x} + x \cdot (-e^{-x}) = e^{-x}(1 - x) (b) Find the coordinates of the stationary point and determine its nature.
d y d x = 0 ⟹ 1 − x = 0 ⟹ x = 1 \dfrac{dy}{dx} = 0 \implies 1 - x = 0 \implies x = 1 d x d y = 0 ⟹ 1 − x = 0 ⟹ x = 1
y = e − 1 = 1 e y = e^{-1} = \dfrac{1}{e} y = e − 1 = e 1
d 2 y d x 2 = − e − x ( 1 − x ) + e − x ( − 1 ) = e − x ( x − 2 ) \dfrac{d^2y}{dx^2} = -e^{-x}(1-x) + e^{-x}(-1) = e^{-x}(x - 2) d x 2 d 2 y = − e − x ( 1 − x ) + e − x ( − 1 ) = e − x ( x − 2 )
At x = 1 x = 1 x = 1 d 2 y d x 2 = e − 1 ( − 1 ) < 0 \dfrac{d^2y}{dx^2} = e^{-1}(-1) \lt 0 d x 2 d 2 y = e − 1 ( − 1 ) < 0 ( 1 , 1 e ) \left(1, \dfrac{1}{e}\right) ( 1 , e 1 )
(c) Find the point of inflection.
d 2 y d x 2 = 0 ⟹ x = 2 \dfrac{d^2y}{dx^2} = 0 \implies x = 2 d x 2 d 2 y = 0 ⟹ x = 2
At x = 2 x = 2 x = 2 y = 2 e − 2 = 2 e 2 y = 2e^{-2} = \dfrac{2}{e^2} y = 2 e − 2 = e 2 2
Since d 2 y d x 2 \dfrac{d^2y}{dx^2} d x 2 d 2 y x = 2 x = 2 x = 2 ( 2 , 2 e 2 ) \left(2, \dfrac{2}{e^2}\right) ( 2 , e 2 2 )
(d) Find the equation of the tangent at x = 0 x = 0 x = 0
y ( 0 ) = 0 y(0) = 0 y ( 0 ) = 0 y ′ ( 0 ) = e 0 ( 1 ) = 1 y'(0) = e^0(1) = 1 y ′ ( 0 ) = e 0 ( 1 ) = 1
Y = x Y = x Y = x Question 5 (Paper 1 style) The radius of a circle is increasing at a rate of 0.5 c m / s 0.5\mathrm{ cm/s} 0.5 cm/s 4 c m 4\mathrm{ cm} 4 cm
A = π r 2 A = \pi r^2 A = π r 2 \frac`\{dA}``\{dt}` = 2\pi r \frac`\{dr}``\{dt}` = 2\pi(4)(0.5) = 4\pi \mathrm{ cm}^2\mathrm{/s} Question 6 (Paper 2 style) Let f ( x ) = x x 2 + 1 f(x) = \dfrac{x}{x^2 + 1} f ( x ) = x 2 + 1 x
(a) Find f ′ ( x ) f'(x) f ′ ( x )
Using the quotient rule with u = x$$v = x^2 + 1 :
F ′ ( x ) = ( x 2 + 1 ) ( 1 ) − x ( 2 x ) ( x 2 + 1 ) 2 = 1 − x 2 ( x 2 + 1 ) 2 F'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{1 - x^2}{(x^2+1)^2} F ′ ( x ) = ( x 2 + 1 ) 2 ( x 2 + 1 ) ( 1 ) − x ( 2 x ) = ( x 2 + 1 ) 2 1 − x 2 (b) Find the range of values of x x x f f f
f ′ ( x ) > 0 ⟹ 1 − x 2 > 0 ⟹ x 2 < 1 ⟹ − 1 < x < 1 f'(x) \gt 0 \implies 1 - x^2 \gt 0 \implies x^2 \lt 1 \implies -1 \lt x \lt 1 f ′ ( x ) > 0 ⟹ 1 − x 2 > 0 ⟹ x 2 < 1 ⟹ − 1 < x < 1
(c) Find the stationary points.
f ′ ( x ) = 0 ⟹ x = ± 1 f'(x) = 0 \implies x = \pm 1 f ′ ( x ) = 0 ⟹ x = ± 1
At x = 1 x = 1 x = 1 f ( 1 ) = 1 2 f(1) = \dfrac{1}{2} f ( 1 ) = 2 1
At x = − 1 x = -1 x = − 1 f ( − 1 ) = − 1 2 f(-1) = -\dfrac{1}{2} f ( − 1 ) = − 2 1
Since f ′ f' f ′ + + + − - − x = 1 x = 1 x = 1 ( 1 , 1 2 ) \left(1, \dfrac{1}{2}\right) ( 1 , 2 1 )
Since f ′ f' f ′ − - − + + + x = − 1 x = -1 x = − 1 ( − 1 , − 1 2 ) \left(-1, -\dfrac{1}{2}\right) ( − 1 , − 2 1 )
(d) Find the equations of the asymptotes.
As x → ± ∞ x \to \pm\infty x → ± ∞ f ( x ) → 0 f(x) \to 0 f ( x ) → 0 y = 0 y = 0 y = 0
The denominator x 2 + 1 ≠ 0 x^2 + 1 \neq 0 x 2 + 1 = 0 x x x
Summary of Key Results Concept Formula First principles f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h \displaystyle f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x ) Power rule d d x [ x n ] = n x n − 1 \dfrac{d}{dx}[x^n] = nx^{n-1} d x d [ x n ] = n x n − 1 Product rule ( u v ) ′ = u ′ v + u v ′ (uv)' = u'v + uv' ( uv ) ′ = u ′ v + u v ′ Quotient rule ( u v ) ′ = u ′ v − u v ′ v 2 \left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2} ( v u ) ′ = v 2 u ′ v − u v ′ Chain rule d y d x = d y d u ⋅ d u d x \dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx} d x d y = d u d y ⋅ d x d u Tangent at ( a , f ( a ) ) (a, f(a)) ( a , f ( a )) y − f ( a ) = f ′ ( a ) ( x − a ) y - f(a) = f'(a)(x - a) y − f ( a ) = f ′ ( a ) ( x − a ) Normal gradient − 1 f ′ ( a ) -\dfrac{1}{f'(a)} − f ′ ( a ) 1 Increasing f ′ ( x ) > 0 f'(x) \gt 0 f ′ ( x ) > 0 Decreasing f ′ ( x ) < 0 f'(x) \lt 0 f ′ ( x ) < 0 Local max f'(a) = 0$$f''(a) \lt 0 Local min f'(a) = 0$$f''(a) \gt 0
For Paper 2 differentiation questions, always show your working . State which rule you are Using (product, quotient, chain) and lay out the substitution. Examiners award method marks for Correct application of rules even if arithmetic errors occur later.
For the A-Level treatment of this topic, see Differentiation .
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Common Pitfalls Forgetting to use the chain rule for composite functions — identify the inner function first.
Confusing the derivative of ln ( x ) \ln(x) ln ( x ) log a ( x ) \log_a(x) log a ( x )
Misreading the question, particularly with ‘hence’ vs ‘hence or otherwise’ — the former requires using previous work.
Summary Chain rule: d d x [ f ( g ( x ) ) ] = f ′ ( g ( x ) ) ⋅ g ′ ( x ) \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) d x d [ f ( g ( x ))] = f ′ ( g ( x )) ⋅ g ′ ( x ) Product rule: ( u v ) ′ = u ′ v + u v ′ (uv)' = u'v + uv' ( uv ) ′ = u ′ v + u v ′ ( u v ) ′ = u ′ v − u v ′ v 2 \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} ( v u ) ′ = v 2 u ′ v − u v ′ Implicit differentiation: apply chain rule to terms with y y y Related rates: differentiate with respect to t t t Cross-References Topic Site Link [Differentiation] A-Level View [Differentiation] IB View
Worked Examples Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.