Derivatives and stationary points · Dérivées et points stationnaires
| English | Français |
|---|---|
| derivative/dɪˈrɪvətɪv/ | dérivée |
What is the slope at one point?
- A curved road has different slopes at different positions. An average gradient cannot describe every point.
- This lesson studies derivative 导数: The instantaneous rate of change, also the gradient of a tangent.
Choose the mathematical structure
- For y=ax^n, dy/dx=anx^(n-1). A stationary point satisfies dy/dx=0. Check the sign change of the derivative, or the second derivative when it is nonzero, to classify it.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines derivative? · Quelle description définit correctement la dérivée ?
The instantaneous rate of change, also the gradient of a tangent. · Le taux de variation instantané, également la pente d'une tangente.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
For y=x³-3x, dy/dx=3x²-3. At x=1, the gradient is 0 and y=-2. The second derivative is 6x, positive at x=1, so this is a local minimum. At x=-1, y=2 and the second derivative is negative, giving a local maximum.
Derivatives and stationary points · Dérivées et points stationnaires
For y=ax^n, dy/dx=anx^(n-1) · Pour y=ax^n, dy/dx=anx^(n-1)
Compare the model with the worked case and explain one change. · Compare le modèle avec l'exemple résolu et explique un changement.
For y=x³-3x, find dy/dx at x=2. · Pour y=x³-3x, trouvez dy/dx en x=2.
dy/dx=3x²-3. At x=2 it is 12-3=9. · dy/dx=3x²-3. En x=2, c'est 12-3=9.
Test a tempting shortcut
- A zero derivative does not always mean a maximum or minimum: y=x³ is stationary at 0 but continues increasing. An endpoint can also produce an extreme value on a restricted domain.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Every point with zero derivative is a local maximum or minimum. This claim is false. Explain which definition or assumption it violates.
Find the y-coordinate of the stationary point at x=1. · Trouvez la coordonnée y du point stationnaire en x=1.
Substitute into the original curve: y(1)=1-3=-2. · Substituez dans la courbe d'origine : y(1)=1-3=-2.
Every point with zero derivative is a local maximum or minimum. · Chaque point avec une dérivée nulle est un maximum ou un minimum local.
A zero derivative does not always mean a maximum or minimum: y=x³ is stationary at 0 but continues increasing. An endpoint can also produce an extreme value on a restricted domain. · Une dérivée nulle ne signifie pas toujours un maximum ou un minimum : y=x³ est stationnaire en 0 mais continue d'augmenter. Une extrémité peut également produire une valeur extrême sur un domaine restreint.
Interpret a new situation
- At GCSE/IGCSE use only the polynomial scope allowed by the tier; do not add chain, product or quotient rules there. At advanced level, connect the derivative to rates and optimization with a valid domain.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
For y=4x³, find dy/dx at x=2. · Pour y=4x³, trouvez dy/dx en x=2.
The derivative is 12x²; at 2 it is 12×4=48. · La dérivée est 12x² ; en 2, c'est 12×4=48.
Match each part of a complete solution to its purpose. · Associer chaque partie d'une solution complète à son but.
An assumption justifies the model; a check tests the result; interpretation connects it to the question. · Une hypothèse justifie le modèle ; une vérification teste le résultat ; l'interprétation le relie à la question.
Use this in your course
- edexcel IAL pure mathematics; official unit P2. Other-unit enrichment is identified in the scope review; it is not an extra cash-in requirement.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
The instantaneous rate of change, also the gradient of a tangent. Choose the relationship, show the method, check its assumptions and interpret the result.