Chain, product, quotient and implicit differentiation · Dérivation par la chaîne, produit, quotient et implicite
| English | Français |
|---|---|
| chain rule/tʃeɪn ruːl/ | règle de composition |
Does the inside expression change too?
- A cost curve is a power of a changing expression. Differentiating the outer power alone misses the rate of its input.
- This lesson studies chain rule 链式法则: The rule that multiplies the outer derivative by the inner derivative for a composite function.
Choose the mathematical structure
- For y=f(g(x)), y prime=f prime(g(x))g prime(x). For uv, differentiate to u prime v+uv prime. For u/v, use (u prime v-uv prime)/v², where v≠0.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines chain rule? · Quelle description définit correctement la règle de la chaîne ?
The rule that multiplies the outer derivative by the inner derivative for a composite function. · La règle qui multiplie la dérivée externe par la dérivée interne pour une fonction composée.
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=(3x+1)^4, y prime=4(3x+1)^3×3=12(3x+1)^3. At x=0, the gradient is 12. For x²+y²=25, differentiate implicitly: 2x+2y y prime=0, so y prime=-x/y when y≠0.
Chain, product, quotient and implicit differentiation · Dérivation par la chaîne, produit, quotient et implicite
For y=f(g(x)), y prime=f prime(g(x))g prime(x) · Pour y=f(g(x)), y prime=f prime(g(x))g prime(x)
Compare the model with the worked case and explain one change. · Compare le modèle avec l'exemple résolu et explique un changement.
Find the derivative of (3x+1)^4 at x=0. · Trouver la dérivée de (3x+1)^4 en x=0.
Chain rule gives 4(3x+1)³×3; at zero this is 12. · La règle de la chaîne donne 4(3x+1)³×3 ; en zéro, cela vaut 12.
Test a tempting shortcut
- A derivative of a product is not the product of the derivatives. In implicit differentiation, every differentiated function of y brings a dy/dx factor. A quotient's denominator is squared.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
The derivative of u(x)v(x) is always u prime times v prime. This claim is false. Explain which definition or assumption it violates.
For x²+y²=25, find dy/dx at (3,4). · Pour x²+y²=25, trouver dy/dx au point (3,4).
Implicit differentiation gives 2x+2y y prime=0, so y prime=-3/4. · La dérivation implicite donne 2x+2y y prime=0, donc y prime=-3/4.
The derivative of u(x)v(x) is always u prime times v prime. · La dérivée de u(x)v(x) est toujours égale à u prime fois v prime.
A derivative of a product is not the product of the derivatives. In implicit differentiation, every differentiated function of y brings a dy/dx factor. A quotient's denominator is squared. · La dérivée d'un produit n'est pas le produit des dérivées. Dans la dérivation implicite, chaque fonction dérivée de y apporte un facteur dy/dx. Le dénominateur d'un quotient est mis au carré.
Interpret a new situation
- Choose a useful form before differentiating: expanding a short polynomial may be simpler. For related rates, write the relation in symbols, differentiate with respect to time, then substitute measured values.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
For y=xe^x, find dy/dx at x=0. · Pour y=xe^x, trouver dy/dx en x=0.
Product rule gives e^x+xe^x; at zero this is 1. · La règle du produit donne e^x+xe^x ; en zéro, cela vaut 1.
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 P4. 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 rule that multiplies the outer derivative by the inner derivative for a composite function. Choose the relationship, show the method, check its assumptions and interpret the result.