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Catégorie :Category: nCreator TI-Nspire
Auteur Author: dissstt
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 3.00 Ko KB
Mis en ligne Uploaded: 20/07/2025 - 21:18:54
Uploadeur Uploader: dissstt (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : https://tipla.net/a4801857
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 3.00 Ko KB
Mis en ligne Uploaded: 20/07/2025 - 21:18:54
Uploadeur Uploader: dissstt (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : https://tipla.net/a4801857
Description
Fichier Nspire généré sur TI-Planet.org.
Compatible OS 3.0 et ultérieurs.
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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)/dx = x/(4 + x^2) 1/x + x/(4 x^2). So: y' = y·[x/(4 + x^2) 1/x + x/(4 x^2)]. 37 y = x^n(a + bx)^m. ln y = n ln x + m ln(a + bx). Differentiate: y'/y = n/x + mb/(a + bx). Thus: y' = x^n(a + bx)^m·[n/x + mb/(a + bx)]. 9.Differentiate d/dx (3x^4 2x^2 + 8). Since d(3x^4)/dx = 12x^3, d(2x^2)/dx = 4x, d(8)/dx = 0, we get: d/dx(3x^4 2x^2 + 8) = 12x^3 4x. Made with nCreator - tiplanet.org
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Compatible OS 3.0 et ultérieurs.
<<
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)/dx = x/(4 + x^2) 1/x + x/(4 x^2). So: y' = y·[x/(4 + x^2) 1/x + x/(4 x^2)]. 37 y = x^n(a + bx)^m. ln y = n ln x + m ln(a + bx). Differentiate: y'/y = n/x + mb/(a + bx). Thus: y' = x^n(a + bx)^m·[n/x + mb/(a + bx)]. 9.Differentiate d/dx (3x^4 2x^2 + 8). Since d(3x^4)/dx = 12x^3, d(2x^2)/dx = 4x, d(8)/dx = 0, we get: d/dx(3x^4 2x^2 + 8) = 12x^3 4x. Made with nCreator - tiplanet.org
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