Advanced Calculus HL

IB Mathematics: Applications & Interpretation · Topic 5: Calculus

JMaths

www.jmaths.xyz

1

Derivatives of Standard Functions

These results are given in the formula booklet. Know them instantly for chain rule applications.

Trigonometric

\( \frac{d}{dx} (\sin x) = \cos x \)
\( \frac{d}{dx} (\cos x) = -\sin x \)
\( \frac{d}{dx} (\tan x) = \sec^2 x \)

Exponential & Logarithmic

\( \frac{d}{dx} (e^x) = e^x \)
\( \frac{d}{dx} (a^x) = a^x \ln a \)
\( \frac{d}{dx} (\ln x) = 1/x \)

✗

Common error: \( \frac{d}{dx} (\cos x) = -\sin x \). The negative sign is the most frequently forgotten detail in calculus exams.

2

Chain Rule, Product Rule & Quotient Rule

Chain rule (composite functions)

\( \frac{d}{dx} [f(g(x))] = f'(g(x)) \times g'(x) \)

“Differentiate the outside, multiply by the derivative of the inside”

Product rule

\( \frac{d}{dx} [uv] = u'v + uv' \)

Quotient rule

\( \frac{d}{dx} [u/v] = (u'v - uv') / v^2 \)

Worked Example — Chain Rule

Differentiate \( y = \) sin(3\( x^2) \).

Let \( u = 3x^2, \) so \( y = \) sin \( u \)

\( dy/dx = \) cos(\( u) \times 6x = 6x \) cos(3\( x^2) \)

Answer: \( dy/dx = 6x \) cos(3\( x^2) \)

Worked Example — Product Rule

Differentiate \( y = x^2 e \)^{3\( x \)}.

\( u = x^2, v = e \)^{3\( x \)} \( \to \) \( u' = 2x, v' = 3e \)^{3\( x \)}

\( dy/dx = 2x e \)^{3\( x \)} + \( x^2 \times 3e \)^{3\( x \)} = \( xe \)^{3\( x \)}(2 + 3\( x) \)

Answer: \( dy/dx = xe \)^{3\( x \)}(2 + 3\( x) \)

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When to use which: Chain rule = function inside a function. Product rule = two functions multiplied. Quotient rule = one function divided by another. Many problems need chain + product together.

3

Second Derivative, Concavity & Inflection

Concave up

\( f''(x) > 0 \)

Curve bends upward (holds water)

Concave down

\( f''(x) < 0 \)

Curve bends downward (spills water)

Point of inflection

\( f''(x) = 0 \) AND concavity changes sign

Check \( f'' \) either side — it must change from + to \( - \) or \( - \) to +

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Common error: \( f''(x) = 0 \) alone does NOT guarantee an inflection point. You must verify the sign change. e.g. \( f(x) = x^4 \) has \( f''(0) = 0 \) but no inflection at \( x = 0 \).

Advanced Calculus HL

IB Mathematics: Applications & Interpretation · Topic 5: Calculus

JMaths

www.jmaths.xyz

4

Integration Techniques

Integration is the reverse of differentiation. These standard results are given in the formula booklet.

Standard integrals

\( \int x^n dx = x^{n+1}/(n+1) + C \)
\( \int e^x dx = e^x + C \)
\( \int 1/x dx = \ln |x| + C \)

Trigonometric integrals

\( \int \cos x dx = \sin x + C \)
\( \int \sin x dx = -\cos x + C \)
\( \int \sec^2 x dx = \tan x + C \)

Reverse chain rule (by inspection)

\( \int f'(g(x)) \times g'(x) dx = f(g(x)) + C \)

e.g. \( \int 6x \) cos(3\( x^2) \) d\( x = \) sin(3\( x^2) + C \) (reverse of the chain rule example above)

✗

Common error: Forgetting the constant of integration \( C \) for indefinite integrals. This loses marks every time. Definite integrals do NOT need + \( C \).

5

Area Between Curves

Area between two curves

\( Area = \int _a^b |f(x) - g(x)| dx \)

If \( f(x) \geq g(x) \) on [\( a, b] \): Area = \( \int \)_{\( a \)}^{\( b \)} [\( f(x) - g(x)] \\, dx \)

Worked Example

Find the area enclosed between \( y = x^2 \) and \( y = x + 2 \).

Intersections: \( x^2 = x + 2 \to x^2 - x - 2 = 0 \to (x - 2)(x + 1) = 0 \to x = -1, 2 \)

Line is above parabola on [\( -1, 2] \)

Area = \( \int _{-1}^2 \) [(\( x + 2) - x^2] \) d\( x = \) [\( x^2/2 + 2x - x^3/3]_{-1}^2 \)

= (2 + 4 \( - 8/3) - (1/2 - 2 + 1/3) = 10/3 - (-7/6) = 9/2 \)

Answer: 4.5 square units

GDC: Integration & Area

TI-84 Plus CE

Graph both functions in Y1 and Y2
[2ND][CALC] → ∫f(x)dx → enter lower & upper bounds
Or: [MATH] → fnInt(Y1\( -Y2, \) X, \( -1, 2) \) for exact numeric answer

TI-Nspire CX II

Graph f1 and f2 → [Menu] → Analyze Graph → Integral
Select function, set bounds → shaded area displayed
Or in Calculator: \( nInt(f1(x)-f2(x), x, -1, 2) \)

Casio fx-CG50

Graph both functions → [SHIFT][G-Solv] → ∫dx
Set lower = \( -1, \) upper = 2
For area between: use \( \int (Y1-Y2)dx \) in Run-Matrix mode

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IB Exam tip: Sketch the curves first to identify which is “on top”. If curves cross within the interval, split into sub-intervals or use absolute value.

Advanced Calculus HL

IB Mathematics: Applications & Interpretation · Topic 5: Calculus

JMaths

www.jmaths.xyz

6

Differential Equations — Separation of Variables

A separable ODE has the form \( dy/dx = f(x) \times g(y). \) Separate the variables and integrate both sides.

Method

\( dy/dx = f(x) g(y) \to \int 1/g(y) dy = \int f(x) dx \)

Worked Example

Solve \( dy/dx = 2xy, \) given \( y(0) = 3 \).

Separate: \( \int 1/y \) d\( y = \int 2x \\, dx \)

ln |y|= x^2 + C \)

\( y = Ae^{x^2} \) where \( A = \) \( e^{C} \)

Initial condition: 3 = \( Ae^0 \to A = 3 \)

Answer: \( y = 3e^{x^2} \)

✗

Common error: Forgetting the constant \( C \) after integrating, then being unable to use the initial condition. Always include + \( C \) on one side before applying the boundary condition.

7

Slope Fields

A slope field (direction field) shows short line segments at grid points with gradient equal to \( dy/dx \) at that point. Solution curves follow the flow of the field.

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Reading slope fields: Where \( dy/dx = 0, \) segments are horizontal. Where \( dy/dx \) is large positive, segments are steep upward. Sketch solution curves by following the arrows smoothly.

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IB Exam tip: You may be given a slope field and asked to sketch a particular solution through a given point. Draw a smooth curve that is tangent to the slope segments at every point.

8

Euler’s Method

A numerical method for approximating solutions to \( dy/dx = f(x, y) \) with step size \( h \).

Euler’s method iteration

\( x_{n+1} = x_n + h y_{n+1} = y_n + h \times f(x_n, y_n) \)

Worked Example

Use Euler’s method with \( h = 0.1 \) to estimate \( y(0.3) \) given \( dy/dx = x + y, y(0) = 1 \).

\( n \)	\( x_{n} \)	\( y_{n} \)	\( f(x_n, y_n) \)	\( h \times f \)
0	0	1	0 + 1 = 1	0.1
1	0.1	1.1	0.1 + 1.1 = 1.2	0.12
2	0.2	1.22	0.2 + 1.22 = 1.42	0.142
3	0.3	1.362	—	—

Answer: \( y(0.3) \approx 1.362 \)

✗

Common error: Using the new \( y \) value to calculate \( f \) in the same step. Always use (\( x_{n} , y_{n} \)) to find the slope, then update to get \( y_{n+1} \).

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Accuracy: Smaller \( h \) gives better accuracy but more steps. Euler’s method always underestimates for concave-up curves and overestimates for concave-down curves.

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Formula booklet: The Euler’s method formula, standard derivatives and integrals, chain/product/quotient rules are all given. The key skill is knowing when and how to apply them.