Soluzioni tutorato3

Manuela Grella e Simona Giovannetti 15 marzo 2005

Esercizio 1. (i)y′ = 5x^{4} − 12x^{2} + 2.

(ii)y = x^{2} \sqrt[3]{x^{2}} = x^{2}x^{2/3} = x^{8/3}, quindi y′ = (8/3)x^{5/3}. (iii)y′ = \frac{2(x^{2}−5x+5)−(2x+3)(2x−5)}{(x^{2}−5x+5)^{2}} = \frac{2x^{2}−10x+10−(4x^{2}−10x+6x−15)}{(x^{2}−5x+5)^{2}} = \frac{−2x^{2}−6x+25}{(x^{2}−5x+5)^{2}}. (iv)y′ = \frac{1}{cos^{2} x} + \frac{1}{sin^{2} x} = \frac{1}{sin^{2} x cos^{2} x}. (v)y′ = \frac{1}{2}(2x arctan x + \frac{1}{1+x^{2}} (1 + x^{2}) − 1) = x arctan x. (vi)y′ = \frac{5x^{4}e^{x}−e^{x}x^{5}}{e^{2x}} = \frac{5x^{4}−x^{5}}{e^{x}}. (vii)y′ = e^{x} arcsin x + e^{x} \frac{1}{\sqrt{1−x^{2}}} . (viii)y′ = \frac{2x ln x− \frac{1}{x} x^{2}}{ln^{2}(x)} = \frac{x(2 ln x−1)}{ln^{2} x}. (ix)y′ = \frac{1}{x} log_{10} x + \frac{ln x}{x} log_{10} e − ln a\frac{1}{x} log_{a} e = \frac{1}{x} log_{10} x + \frac{ln x}{x ln 10} − \frac{1}{x} =

2\frac{ln x}{x ln 10} − \frac{1}{x} dove abbiamo usato il fatto che log_{a} b = \frac{ln b}{ln a}. (x)y′ = \frac{1}{3}(sin^{2} x)^{− \frac{2}{3}} 2 sin x cos x + 3 cos^{−4} x sin x = \frac{2}{3} \frac{cos x}{\sqrt[3]{sin x}} + 3\frac{sin x}{cos^{4} x}. (xi)y′ = 3 cos(3x) − \frac{1}{5} sin \frac{x}{5} + \frac{1}{2} \frac{1}{\sqrt{x} cos^{2}(\sqrt{x})}. (xii)y′ = \frac{3}{4} \frac{x^{2}−1}{x^{2}+1} \frac{2x(x^{2}−1)−2x(x^{2}+1)}{(x^{2}−1)^{2}} + \frac{1}{4} \frac{x+1}{x−1} \frac{x+1−x+1}{(x+1)^{2}} + \frac{1}{2} \frac{1}{1+x^{2}} = \frac{−3x}{(x^{2}+1)(x^{2}−1)} +

\frac{1}{2} \frac{1}{(x−1)(x+1)} + \frac{1}{2} \frac{1}{x^{2}+1} = \frac{−6x+x^{2}+1+x^{2}−1}{2(x^{2}+1)(x+1)(x−1)} = \frac{x^{2}−3x}{(x^{4}−1)} (xiii)y′ = \frac{e^{x}−e^{−x}}{2} = sh(x) (xiv)y′ = 3^{x}log3 (xv)y′ = 2cosx (xvi)y′ = cos(x^{2} + 1)2x (xvii)y′ = \frac{1}{x^{2}−5}2x (xviii)y′ = e^{cos2x+sen2x}(−2sen2x + 2cos2x) (xix)y′ = sh(\sqrt{senx})\frac{1}{2}senx^{1/2−1}cosx = \frac{sh(\sqrt{senx})cosx}{2\sqrt{senx}} (xx)y′ = \frac{2x^{3}senx−2xe^{x}senx+x^{4}cosx−x^{2}e^{x}cosx−2x^{3}e^{x}senx−x^{4}e^{x}senx+x^{2}e^{x}senx}{(x^{2}−1)^{2}e^{2x}} =

\frac{x^{4}(cosx−e^{x}senx)+2x^{3}(senx−e^{x}senx)+x^{2}(e^{x}senx−e^{x}cosx)−2xe^{x}senx}{(x^{2}−1)^{2}e^{2x}}

(xxi)y′ = \frac{2x(x^{2}+1)−2x(x^{2}−1)}{(x^{2}+1)^{2}} = \frac{4x}{(x^{2}+1)^{2}} (xxii)y′ = \frac{1}{x−\sqrt{x^{2}−3}}(1 − \frac{x}{\sqrt{x^{2}−3}}) (xxiii)y′ = 6cosx − \frac{1}{senx+\sqrt{sen^{2}x−1/4}}(cosx + \frac{2senxcosx}{2\sqrt{sen^{2}x−1/4}}) (xxiv)se x > 0 y = e^{arctan( \frac{1+x}{1−x})−1/2ln(1+x^{2})} e y′ = e^{arctan( \frac{1+x}{1−x})−1/2ln(1+x^{2})}(\frac{1−x}{1+x^{2}} ),se

x < 0 y = e^{π/4−ln(1−x)} e y′ = \frac{e^{π/4−ln(1−x)}}{1−x}

Esercizio 2. lim_{h→0} \frac{sen(x+h)−senx}{h} = lim_{h→0}2\frac{sen(h/2)cos( \frac{2x+h}{2} )}{h} = lim_{h→0}cos(\frac{2x+h}{2}) = cosx. Esercizio 3. La pendenza di una curva é data dalla tangente dell’angolo formato dalla tangente alla curva nel punto in questione;quindi dalla derivata della funzione calcolata in quel punto;in questo caso la derivata é y′ = x/2,che,calcolata nel punto é 1,a cui corrisponde un angolo di π/4. Esercizio 4. (i)y′ = \frac{6a}{\sqrt{a^{2}+b^{2}}} x^{5}.

(ii)y′ = −\frac{2}{3}ax^{− \frac{2}{3} −1} + b \frac{\sqrt[3]{x}+ \frac{1}{3} x^{1− \frac{2}{3}}}{x^{2} \sqrt[3]{x^{2}}} = −\frac{2}{3}ax^{− \frac{5}{3}} + \frac{4}{3}bx^{− \frac{7}{3}} = \frac{4b}{3x^{2} \sqrt[3]{x}} − \frac{2a}{3x \sqrt[3]{x^{2}}} . (iii)y′ = \frac{(cos x−sin x)(sin x−cos x)−(sin x+cos x)(cos x+sin x)}{(sin x−cos x)^{2}} = −\frac{2}{(sin x−cos x)^{2}} . (iv)y′ = 7x^{6}e^{x} + x^{7}e^{x}. (v)y′ = −\frac{1}{x^{2}} + 2\frac{1}{x} − \frac{1−ln x}{x^{2}} = \frac{2}{x} − \frac{2}{x^{2}} + \frac{ln x}{x^{2}}. (vi)y′ = 30(1 + 3x − 5x^{2})^{29}(3 − 10x). (vii)y′ = (2x − 5) cos(x^{2} − 5x + 1) − \frac{a}{x^{2}} \frac{1}{cos^{2} \frac{a}{x}}.

(viii)y′ = \frac{4}{3}(\frac{x−1}{x+2})^{− \frac{3}{4}} \frac{x+2−x+1}{(x+2)^{2}} = 4(x − 1)^{− \frac{3}{4}} (x + 2)^{− \frac{5}{4}} .

(ix)y′ = 3cosx + 5senx

(x)y′ = \frac{2x}{x^{2}−5}

(xi)y′ = \frac{log_{3}ecosx}{senx}

(xii)y′ = ^{\frac{2^{x}\sqrt{x}shxlog2log\sqrt{x}+ \frac{2^{x}shx}{2\sqrt{x}} − \frac{2^{x}logxshx}{2\sqrt{x}} −2^{x}logx\sqrt{x}chx}{(shx)^{2}x}}

(xiii)y′ = −^{\frac{e^{x}log(\sqrt{x^{2}+1})+ \frac{e^{x}x}{x^{2}+1}}{e^{2x}log^{2}(\sqrt{x^{2}+1})}}