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.
GDC: Derivative at a Point
TI-84 Plus CE
Store the function in Y1 [MATH] → nDeriv( → \( \text{nDeriv}(Y_1, X, a) \) gives \( f'(a) \)
TI-Nspire CX II
[Menu] → Calculus → Numerical Derivative
Enter the \( \tfrac{d}{dx} \) template at \( x = a \) (or on a Graph: Analyze Graph → dy/dx)
Casio fx-CG50
Graph the function → [SHIFT][G-Solv] → dy/dx
Enter \( x = a \) to read off the gradient \( f'(a) \)
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 \( x = 0 \).
4
Related Rates of Change
When two quantities both vary with time, the chain rule links their rates.
Method: Write the formula connecting the quantities, differentiate to get \( dV/dr \) (or similar), then multiply by the given rate. Substitute the instant's values only at the end.
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. \( \int 6x \cos(3x^2) \, dx = \sin(3x^2) + C \) (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 + \( C \).
Graph both functions in Y1 and Y2 [2ND][CALC] → ∫f(x)dx → enter lower & upper bounds
Or: [MATH] → \( \text{fnInt}(Y_1 - Y_2, X, -1, 2) \) 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: \( \text{nInt}(f_1(x) - f_2(x), x, -1, 2) \)
Casio fx-CG50
Graph both functions → [SHIFT][G-Solv] → ∫dx
Set lower = \( -1, \) upper = \( 2 \)
For area between: use \( \int (Y_1 - Y_2)\,dx \) 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.
Answer: displacement \( = -9 \) m (use \( \int_0^3 |v|\,dt \) on the GDC for total distance)
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IB Exam tip: “Total distance travelled” almost always needs \( \int |v|\,dt \). Find where \( v = 0 \) first, then integrate \( |v| \) on the GDC across the whole interval.
Setting up a DE from words: translate the context into a rate. “Rate of change of \( N \) proportional to \( N \)” \( \to \dfrac{dN}{dt} = kN \); “tank drains at a rate proportional to volume” \( \to \dfrac{dV}{dt} = -kV \). Then separate and solve.
Worked Example
Solve \( dy/dx = 2xy, \) given \( y(0) = 3 \).
Separate: \( \int 1/y \, dy = \int 2x \, dx \)
\( \ln|y| = x^2 + C \)
\( y = Ae^{x^2} \) 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.
10
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.
A numerical method for approximating solutions to \( dy/dx = f(x, y) \) with step size \( h \).
Euler’s method iteration
\( x_{n+1} = x_n + h \) \( y_{n+1} = y_n + h \, f(x_n, y_n) \)
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.
A pair of linked rates \( \dfrac{dx}{dt} = ax + by, \; \dfrac{dy}{dt} = cx + dy \) is written as the matrix \( M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \). The eigenvalues of \( M \) determine the long-term behaviour.
Both real & positive → unstable (source). Both real & negative → stable node (sink). Opposite signs → saddle. Complex \( \to \) spiral (centre if purely imaginary).
Equilibrium point
\( \dfrac{dx}{dt} = \dfrac{dy}{dt} = 0 \)
Solve for \( (x, y) \) where both rates vanish
Trajectories
Follow eigenvector directions
Real eigenvalues → straight-line solutions along eigenvectors
Worked Example — Classifying a System
Classify \( \dfrac{dx}{dt} = x, \; \dfrac{dy}{dt} = -2y \) at the origin.
\( M = \begin{pmatrix} 1 & 0 \\ 0 & -2 \end{pmatrix} \), diagonal so eigenvalues are \( \lambda = 1 \) and \( \lambda = -2 \)
Real eigenvalues of opposite sign
Answer: the origin is a saddle point (unstable)
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Numerical trajectories: Euler’s method extends to coupled systems — update \( x \) and \( y \) together each step: \( x_{n+1} = x_n + h\,(ax_n + by_n) \) and \( y_{n+1} = y_n + h\,(cx_n + dy_n) \).
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Second-order DEs: a \( \dfrac{d^2 x}{dt^2} \) equation reduces to a coupled first-order system. Let \( v = \dfrac{dx}{dt} \), then \( \dfrac{dx}{dt} = v \) and \( \dfrac{dv}{dt} \) comes from the original equation — now apply Euler to both.
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Common error: updating \( y \) using the new \( x \) within the same Euler step. Compute both increments from the old \( (x_n, y_n) \) values, then update together.
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