Calcolo di derivate

Manuela Grella e Simona Giovannetti

8 marzo 2005

Esercizio 1. Calcolare le seguenti derivate: (i)y = x^{5} − 4x^{3} + 2x − 3 (ii)y = x^{2} \sqrt[3]{x^{2}} (iii)y = \frac{2x+3}{x^{2}−5x+5} (iv)y = tan x − cot x (v)y = \frac{(1+x^{2}) arctan x−x}{2} (vi)y = \frac{x^{5}}{e^{x}} (vii)y = e^{x} arcsin x (viii)y = \frac{x^{2}}{ln x} (ix)y = ln x log_{10} x − ln a log_{a} x (x)y = \sqrt[3]{sin^{2}x} + \frac{1}{cos^{3}x} (xi)y = sin 3x + cos \frac{x}{5} + tan \sqrt{x} (xii)y = \frac{3}{4} ln \frac{x^{2}+1}{x^{2}−1} + \frac{1}{4} ln \frac{x−1}{x+1} + \frac{1}{2} arctan x (xiii)y = ch x = \frac{e^{x}+e^{−x}}{2} (xiv)y = 3^{x} (xv)y = 2 sin x − 1 (xvi)y = sin(x^{2} + 1) (xvii)y = ln(x^{2} − 5) (xviii)y = e^{cos 2x+sin 2x} (xix)y = ch(\sqrt{sin x}) (xx)y = \frac{x^{2} sin x}{(x^{2}−1)e^{x}}

(xxi)y = \frac{x^{2}−1}{x^{2}+1} (xxii)y = ln(x − \sqrt{x^{2} − 3}) (xxiii)y = 6 sin x − ln(sin x + \sqrt{sin^{2} x − \frac{1}{4}})

(xxiv)y = e^{arctan \frac{1+x}{1−|x|} − \frac{1}{2} ln(1+|x|−x+x^{2})}

Esercizio 2. Calcolare usando la definizione la derivata di y = sin x.

Esercizio 3. Calcolare la pendenza della curva y = \frac{1}{4}x^{2} nel punto di ascisse x = 2.

Esercizi per casa

Esercizio 4. Calcolare le seguenti derivate: (i)y = \frac{ax^{6}+b}{\sqrt{a^{2}+b^{2}}} (ii)y = \frac{a}{\sqrt[3]{x^{2}}} − \frac{b}{x \sqrt[3]{x}} (iii)y = \frac{sin x+cos x}{sin x−cos x} (iv)y = x^{7}e^{x} (v)y = \frac{1}{x} + 2 ln x − \frac{ln x}{x} (vi)y = (1 + 3x − 5x^{2})^{30} (vii)y = sin(x^{2} − 5x + 1) + tan \frac{a}{x}

(viii)y = \frac{4}{3} \sqrt[4]{\frac{x−1}{x+2}} (ix)y = 3 sin x − 5 cos x (x)y = ln(tan x + x) (xi)y = log_{3} sin(3x) (xii)y = \frac{2^{x} ln \sqrt{x}}{sh(x)\sqrt{x}} (xiii)y = \frac{1}{e^{x} ln(\sqrt{x^{2}+1})}