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Catégorie :Category: nCreator TI-Nspire
Auteur Author: dissstt
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 2.68 Ko KB
Mis en ligne Uploaded: 20/07/2025 - 21:18:48
Uploadeur Uploader: dissstt (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : https://tipla.net/a4801856
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 2.68 Ko KB
Mis en ligne Uploaded: 20/07/2025 - 21:18:48
Uploadeur Uploader: dissstt (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : https://tipla.net/a4801856
Description
Fichier Nspire généré sur TI-Planet.org.
Compatible OS 3.0 et ultérieurs.
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1. y = ln(ax^2 + b) Differentiate: y' = 1/(ax^2 + b) · 2ax = (2ax)/(ax^2 + b) 3. y = ln((ax + b)^2) Notice that: ln((ax + b)^2) = 2ln(ax + b) Then: y' = 2 · 1/(ax + b) · a = 2a/(ax + b) 4. y = ln(ax^n) We can write: ln(ax^n) = ln a + n ln x So: y' = 0 + n/x = n/x Alternatively, using the chain rule directly: y' = 1/(ax^n) · (a n x^{n-1}) = (anx^{n-1})/(a x^n) = n/x 5. y = ln(x^3) Since ln(x^3) = 3ln x, we obtain: y' = 3 · 1/x = 3/x 6. y = ln^3 x (i.e. y = (ln x)^3) Apply the chain rule: y' = 3(ln x)^2 · 1/x = (3(ln x)^2)/x 7. y = ln(2x^3 3x^2 + 4) Differentiate: y' = 1/(2x^3 3x^2 + 4) · (6x^2 6x) = (6x^2 6x)/(2x^3 3x^2 + 4) 8. y = log(2/x) Assume log means base 10: log(2/x) = log 2 log x Since d/dx log x = 1/(x ln 10), we have: y' = 1/(x ln 10) Alternatively, using the chain rule: y' = 1/(2/x ln 10) · (2/x^2) = 2/x^2 · x/(2 ln 10) = 1/(x ln 10) 9. y = ln(x^2 / (1 + x^2)) Write this as a difference: y = ln x^2 ln(1 + x^2) = 2ln x ln(1 + x^2) Then: y' = 2/x 1/(1 + x^2) · 2x = 2/x 2x/(1 + x^2) 10. y = ln((9 2x^2)) Since (9 2x^2) = (9 2x^2)^{1/2}, write: y = (1/2)ln(9 2x^2) Then: y' = (1/2) · 1/(9 2x^2) · (4x) = 2x/(9 2x^2) 11. y = ln(ax(a + x)) Write the argument as a product: ax(a + x) = ax(a + x)^{1/2} Then: y = ln a + ln x + (1/2)ln(a + x) Differentiate: y' = 0 + 1/x + (1/2) · 1/(a + x) = 1/x + 1/(2(a + x)) 12. f(x) = x ln x Using the product rule: f'(x) = 1·ln x + x·1/x = ln x + 1 13. f(x) = ln(x + (1 + x^2)) Differentiate using the chain rule: f'(x) = 1/(x + (1 + x^2)) · (1 + x/(1 + x^2)) Notice: 1 + x/(1 + x^2) = ((1 + x^2) + x)/(1 + x^2) Then: f'(x) = 1/(1 + x^2) 14. s = ln(((a + bt)/(a bt))) Write: s = (1/2)ln((a + bt)/(a bt)) = (1/2)[ln(a + bt) ln(a bt)] Differentiate: s' = (1/2)[b/(a + bt) + b/(a bt)] = (b/2)(1/(a + bt) + 1/(a bt)) 15. f(x) = x^2 ln(x^2) Notice: ln x^2 = 2 ln|x| So: f(x) = 2x^2 ln|x| Differentiate: f'(x) = 2(2x ln|x| + x^2·1/x) = 4x ln|x| + 2x = 2x(2 ln|x| + 1) Alternatively: d/dx ln x^2 = 2/x Then: f'(x) = 2x ln x^2 + x^2·(2/x) = 2x ln x^2 + 2x 16. y = e^{nx} Differentiate: y' = e^{nx}·n = n e^{nx} 17. y = 10^{nx} Recall: d/dx a^{u(x)} = a^{u(x)} ln a · u'(x) Thus: y' = 10^{nx} ln(10)·n = n ln(10) 10^{nx} 18. y = e^{x^2} Differentiate: y' = e^{x^2}·(2x) = 2x e^{x^2} 19. y = 2/e^x = 2e^{-x} Differentiate: y' = 2·(e^{-x}) = 2e^{-x} = 2/e^x 20. s = e^{vT} Assuming v is constant and differentiating w.r.t. T: ds/dT = e^{vT}·v = v e^{vT} Made with nCreator - tiplanet.org
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Compatible OS 3.0 et ultérieurs.
<<
1. y = ln(ax^2 + b) Differentiate: y' = 1/(ax^2 + b) · 2ax = (2ax)/(ax^2 + b) 3. y = ln((ax + b)^2) Notice that: ln((ax + b)^2) = 2ln(ax + b) Then: y' = 2 · 1/(ax + b) · a = 2a/(ax + b) 4. y = ln(ax^n) We can write: ln(ax^n) = ln a + n ln x So: y' = 0 + n/x = n/x Alternatively, using the chain rule directly: y' = 1/(ax^n) · (a n x^{n-1}) = (anx^{n-1})/(a x^n) = n/x 5. y = ln(x^3) Since ln(x^3) = 3ln x, we obtain: y' = 3 · 1/x = 3/x 6. y = ln^3 x (i.e. y = (ln x)^3) Apply the chain rule: y' = 3(ln x)^2 · 1/x = (3(ln x)^2)/x 7. y = ln(2x^3 3x^2 + 4) Differentiate: y' = 1/(2x^3 3x^2 + 4) · (6x^2 6x) = (6x^2 6x)/(2x^3 3x^2 + 4) 8. y = log(2/x) Assume log means base 10: log(2/x) = log 2 log x Since d/dx log x = 1/(x ln 10), we have: y' = 1/(x ln 10) Alternatively, using the chain rule: y' = 1/(2/x ln 10) · (2/x^2) = 2/x^2 · x/(2 ln 10) = 1/(x ln 10) 9. y = ln(x^2 / (1 + x^2)) Write this as a difference: y = ln x^2 ln(1 + x^2) = 2ln x ln(1 + x^2) Then: y' = 2/x 1/(1 + x^2) · 2x = 2/x 2x/(1 + x^2) 10. y = ln((9 2x^2)) Since (9 2x^2) = (9 2x^2)^{1/2}, write: y = (1/2)ln(9 2x^2) Then: y' = (1/2) · 1/(9 2x^2) · (4x) = 2x/(9 2x^2) 11. y = ln(ax(a + x)) Write the argument as a product: ax(a + x) = ax(a + x)^{1/2} Then: y = ln a + ln x + (1/2)ln(a + x) Differentiate: y' = 0 + 1/x + (1/2) · 1/(a + x) = 1/x + 1/(2(a + x)) 12. f(x) = x ln x Using the product rule: f'(x) = 1·ln x + x·1/x = ln x + 1 13. f(x) = ln(x + (1 + x^2)) Differentiate using the chain rule: f'(x) = 1/(x + (1 + x^2)) · (1 + x/(1 + x^2)) Notice: 1 + x/(1 + x^2) = ((1 + x^2) + x)/(1 + x^2) Then: f'(x) = 1/(1 + x^2) 14. s = ln(((a + bt)/(a bt))) Write: s = (1/2)ln((a + bt)/(a bt)) = (1/2)[ln(a + bt) ln(a bt)] Differentiate: s' = (1/2)[b/(a + bt) + b/(a bt)] = (b/2)(1/(a + bt) + 1/(a bt)) 15. f(x) = x^2 ln(x^2) Notice: ln x^2 = 2 ln|x| So: f(x) = 2x^2 ln|x| Differentiate: f'(x) = 2(2x ln|x| + x^2·1/x) = 4x ln|x| + 2x = 2x(2 ln|x| + 1) Alternatively: d/dx ln x^2 = 2/x Then: f'(x) = 2x ln x^2 + x^2·(2/x) = 2x ln x^2 + 2x 16. y = e^{nx} Differentiate: y' = e^{nx}·n = n e^{nx} 17. y = 10^{nx} Recall: d/dx a^{u(x)} = a^{u(x)} ln a · u'(x) Thus: y' = 10^{nx} ln(10)·n = n ln(10) 10^{nx} 18. y = e^{x^2} Differentiate: y' = e^{x^2}·(2x) = 2x e^{x^2} 19. y = 2/e^x = 2e^{-x} Differentiate: y' = 2·(e^{-x}) = 2e^{-x} = 2/e^x 20. s = e^{vT} Assuming v is constant and differentiating w.r.t. T: ds/dT = e^{vT}·v = v e^{vT} Made with nCreator - tiplanet.org
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