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Catégorie :Category: nCreator TI-Nspire
Auteur Author: dissstt
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 7.90 Ko KB
Mis en ligne Uploaded: 20/07/2025 - 20:56:54
Uploadeur Uploader: dissstt (Profil)
Téléchargements Downloads: 0
Visibilité Visibility: Archive publique
Shortlink : https://tipla.net/a4801838
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 7.90 Ko KB
Mis en ligne Uploaded: 20/07/2025 - 20:56:54
Uploadeur Uploader: dissstt (Profil)
Téléchargements Downloads: 0
Visibilité Visibility: Archive publique
Shortlink : https://tipla.net/a4801838
Description
Fichier Nspire généré sur TI-Planet.org.
Compatible OS 3.0 et ultérieurs.
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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} 21 Given z = b^{2v}. Write it in the form a^{u} with constant base b and exponent 2v. The derivative is dz/dv = b^{2v}·ln b·(2) = 2 ln(b) b^{2v}. 22 Given u = s e^s. Using the product rule: du/ds = d(s)/ds·e^s + s·d(e^s)/ds = e^s + s·e^s = e^s(1+s). 23 Given v = e^u / u. Let N(u)=e^u, D(u)=u, N'(u)=e^u, D'(u)=1. Then: dv/du = [u·e^u e^u] / u^2 = e^u(u1) / u^2. 24 Given y = (ln x) / x. Using quotient rule (or rewriting as ln x·x{¹): y' = [x·(1/x) ln x·1] / x^2 = (1 ln x) / x^2. 25 Given y = ln(x^2 e^x). Use log properties: y = ln(x^2) + ln(e^x) = 2 ln x + x. Differentiate: y' = 2/x + 1. 26 Given y = (e^x 1) / (e^x + 1). Let N=e^x1, D=e^x+1, N'=e^x, D'=e^x. Then numerator simplifies to 2e^x, giving: y' = 2e^x / (e^x + 1)^2. 27 Given y = x^2 e^(x). Using product rule with u=x^2, v=e^(x), v'=e^(x): y' = 2x e^(x) + x^2 (e^(x)) = e^(x)(2x x^2). 28 Given y = (a/2)[ e^(x/a) e^(x/a) ]. Differentiate termbyterm (d/dx e^(±x/a) = ±(1/a)e^(±x/a)): y' = ½(e^(x/a)+e^(x/a)). 29 Given y = (e^x e^(x)) / (e^x + e^(x)), i.e., tanh x. Its derivative is sech²x = 4 / (e^x + e^(x))^2. 30 Given s = ln(t^2) / t^2. Since ln(t^2) = 2 ln|t|, write s = 2ln|t| / t^2. Apply quotient rule: ds/dt = [t^2·(2/t) 2ln|t|·2t] / t^4 = 2(1 2ln|t|) / t^3. 31 Given f(x) = ln[((x^2+1) x)/((x^2+1) + x)]. One shows f(x) = 2 arcsinh(x), and f'(x) = 2 / (x^2+1). 32 Find the derivative of y = x^x. Take ln: ln y = x ln x. Differentiate: y'/y = ln x + 1. Thus: y' = x^x(ln x + 1). 33 y = x^(x). Write ln y = x ln x. d/dx(x ln x) = (1/(2x)) ln x + 1/x. So y'/y = [ln x + 2] / [2x]. Thus: y' = x^(x)·[(ln x + 2)/(2x)]. 34 s = (a/t)^t. ln s = t(ln a ln t). d/dt: s'/s = ln a ln t 1. Hence: s' = (a/t)^t·(ln(a/t) 1). 35 y = x [3]{3x + a} / (2x + b). Rewrite: y = x·(3x + a)^(1/3)(2x + b)^(-1/2). ln y = ln x + (1/3)ln(3x + a) (1/2)ln(2x + b). Differentiate: y'/y = 1/x + 1/(3x + a) 1/(2x + b). So: y' = y·[1/x + 1/(3x + a) 1/(2x + b)]. 36.Problem 36 y = (4 + x^2) / [x (4 x^2)]. Rewrite: y = (4 + x^2)^(1/2)x^(1)(4 x^2)^(1/2). ln y = ½ln(4 + x^2) ln x ½ln(4 x^2). Differentiate: (d ln y)
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Compatible OS 3.0 et ultérieurs.
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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} 21 Given z = b^{2v}. Write it in the form a^{u} with constant base b and exponent 2v. The derivative is dz/dv = b^{2v}·ln b·(2) = 2 ln(b) b^{2v}. 22 Given u = s e^s. Using the product rule: du/ds = d(s)/ds·e^s + s·d(e^s)/ds = e^s + s·e^s = e^s(1+s). 23 Given v = e^u / u. Let N(u)=e^u, D(u)=u, N'(u)=e^u, D'(u)=1. Then: dv/du = [u·e^u e^u] / u^2 = e^u(u1) / u^2. 24 Given y = (ln x) / x. Using quotient rule (or rewriting as ln x·x{¹): y' = [x·(1/x) ln x·1] / x^2 = (1 ln x) / x^2. 25 Given y = ln(x^2 e^x). Use log properties: y = ln(x^2) + ln(e^x) = 2 ln x + x. Differentiate: y' = 2/x + 1. 26 Given y = (e^x 1) / (e^x + 1). Let N=e^x1, D=e^x+1, N'=e^x, D'=e^x. Then numerator simplifies to 2e^x, giving: y' = 2e^x / (e^x + 1)^2. 27 Given y = x^2 e^(x). Using product rule with u=x^2, v=e^(x), v'=e^(x): y' = 2x e^(x) + x^2 (e^(x)) = e^(x)(2x x^2). 28 Given y = (a/2)[ e^(x/a) e^(x/a) ]. Differentiate termbyterm (d/dx e^(±x/a) = ±(1/a)e^(±x/a)): y' = ½(e^(x/a)+e^(x/a)). 29 Given y = (e^x e^(x)) / (e^x + e^(x)), i.e., tanh x. Its derivative is sech²x = 4 / (e^x + e^(x))^2. 30 Given s = ln(t^2) / t^2. Since ln(t^2) = 2 ln|t|, write s = 2ln|t| / t^2. Apply quotient rule: ds/dt = [t^2·(2/t) 2ln|t|·2t] / t^4 = 2(1 2ln|t|) / t^3. 31 Given f(x) = ln[((x^2+1) x)/((x^2+1) + x)]. One shows f(x) = 2 arcsinh(x), and f'(x) = 2 / (x^2+1). 32 Find the derivative of y = x^x. Take ln: ln y = x ln x. Differentiate: y'/y = ln x + 1. Thus: y' = x^x(ln x + 1). 33 y = x^(x). Write ln y = x ln x. d/dx(x ln x) = (1/(2x)) ln x + 1/x. So y'/y = [ln x + 2] / [2x]. Thus: y' = x^(x)·[(ln x + 2)/(2x)]. 34 s = (a/t)^t. ln s = t(ln a ln t). d/dt: s'/s = ln a ln t 1. Hence: s' = (a/t)^t·(ln(a/t) 1). 35 y = x [3]{3x + a} / (2x + b). Rewrite: y = x·(3x + a)^(1/3)(2x + b)^(-1/2). ln y = ln x + (1/3)ln(3x + a) (1/2)ln(2x + b). Differentiate: y'/y = 1/x + 1/(3x + a) 1/(2x + b). So: y' = y·[1/x + 1/(3x + a) 1/(2x + b)]. 36.Problem 36 y = (4 + x^2) / [x (4 x^2)]. Rewrite: y = (4 + x^2)^(1/2)x^(1)(4 x^2)^(1/2). ln y = ½ln(4 + x^2) ln x ½ln(4 x^2). Differentiate: (d ln y)
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