Differentiation: Rules & Applications
Differentiation measures the instantaneous rate of change of a function and gives the gradient of the tangent to a curve at any point. At A-Level Maths, extending differentiation with the chain rule, product rule, quotient rule, and implicit differentiation lets you locate stationary points and solve optimisation problems.
- Chain Rule
- A formula to compute the derivative of a composite function: dy/dx = (dy/du) × (du/dx).
- Product Rule
- A rule used to find the derivative of the product of two functions: d/dx (uv) = u(dv/dx) + v(du/dx).
- Quotient Rule
- A rule for differentiating the quotient of two functions: d/dx (u/v) = (v(du/dx) - u(dv/dx)) / v².
Definite Integral & Trapezium Rule Step-by-Step Visualizer
Choose a function, slide the limits a and b to see the area under the curve update live, and compare Exact Integration vs the Trapezium Rule.
Step-by-Step Working for ∫03 (1x²) dx
- Step 1, Find the antiderivative F(x):F(x) = [(1/3)x³] from x = 0 to x = 3
- Step 2: Substitute upper limit (b = 3) minus lower limit (a = 0):F(3) - F(0) = (9.0000) - (0.0000) = 9.0000
- Step 3, Compare with Trapezium Rule (4 strips, h = 0.750):Area ≈ (h/2)[y₀ + yₙ + 2(y₁ + … + yₙ₋₁)] = 9.2813
Difference from exact integral: 3.13% (increase
nstrips to make the approximation more accurate!).
+ C on indefinite integrals, and use radians whenever integrating trigonometric functions like sin(x) or cos(x)!Key Concepts & Exam Definitions (5)
- Chain Rule
- A formula to compute the derivative of a composite function: dy/dx = (dy/du) × (du/dx).
- Product Rule
- A rule used to find the derivative of the product of two functions: d/dx (uv) = u(dv/dx) + v(du/dx).
- Quotient Rule
- A rule for differentiating the quotient of two functions: d/dx (u/v) = (v(du/dx) - u(dv/dx)) / v².
- Stationary Point
- A point on a curve where the gradient is zero (dy/dx = 0), representing a local maximum, minimum, or point of inflection.
- Parametric Differentiation
- Finding dy/dx when x and y are given in terms of a third variable (parameter) t: dy/dx = (dy/dt) / (dx/dt).
Quick-Fire Flashcards (5)
Click any card to check the answer, then mark it as Known to track your progress.
What is the derivative of e^(kx)?
State the Product Rule.
What is the derivative of ln(x)?
State the Quotient Rule.
How do you test for a local maximum using the second derivative?
Exam-Style Questions & Mark Schemes (3)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Differentiate y = x² e^x with respect to x.
Click “Check Mark Scheme” to compare your answer, Using the product rule (u=x², v=e^x): dy/dx = x² e^x + 2x e^x.
- Q2
Find dy/dx for y = sin(3x²).
Click “Check Mark Scheme” to compare your answer, Using the chain rule: dy/dx = 6x cos(3x²).
- Q3
Determine the nature of the stationary point on y = x² - 4x + 3 at x = 2.
Click “Check Mark Scheme” to compare your answer, dy/dx = 2x - 4. At x=2, dy/dx = 0. d²y/dx² = 2 (which is > 0), so it is a local minimum.
Common Exam Questions (3)
What is the chain rule used for?
The chain rule differentiates composite functions of the form y = f(g(x)) by multiplying the derivative of the outer function by the derivative of the inner function: dy/dx = (dy/du) × (du/dx).
How do I know whether to use the product or quotient rule?
Use the product rule, d/dx(uv) = u(dv/dx) + v(du/dx), when two functions of x are multiplied together. Use the quotient rule, d/dx(u/v) = (v(du/dx) - u(dv/dx)) / v², when one function of x is divided by another.
What does the second derivative tell us?
Evaluating the second derivative d²y/dx² at a stationary point (where dy/dx = 0) determines its nature: if d²y/dx² > 0 the point is a local minimum, and if d²y/dx² < 0 it is a local maximum.
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