Lista 2 - 2023-2

C´alculo 2 - Estat´ıstica Prof. Carlos Freire Segunda Lista Entrega : 16/01/2024 Discente: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A derivada de uma fun¸c˜ao ´e definida da seguinte forma: f ′(x) = lim h−→0 f(x + h) − f(x) h (1) Utilize a defini¸c˜ao de derivada para calcular as fun¸c˜oes abaixo. 1. sen(x). 2. cos(x). 3. tan(x). 4. cossec(x). 5. sec(x). 6. cotg(x). Calcule as seguintes derivadas de fun¸c˜oes trigonom´etricas inversas. 1. (sen−1x). 2. (cos−1 x). 3. (tan−1 x). 4. (cossec−1x). 5. (sec−1 x). 6. (cotg−1x). Exercícios de Cálculo 1) f(x) = sen(x) f'(x) = lim (sen(x+h) - sen(x)) / h h->0 = lim (sen(x)cos(h) + cos(x)sen(h) - sen(x)) / h h->0 = lim (sen(x)cos(0) + cos(x)sen(h) - sen(x)) / h h->0 = lim cos(x) sen(h) / h h->0 f'(x) = cos(x) 2) f(x) = cos(x) f'(x) = lim (cos(x+h) - cos(x)) / h h->0 = lim (cos(x)cos(h) - sen(x)sen(h) - cos(x)) / h h->0 = lim -sen(x) sen(h) / h h->0 f'(x) = -sen(x) 3) f(x) = tg(x) f'(x) = lim (tg(x+h) - tg(x)) / h h->0 = lim (sen(x)/cos(x) - sen(x)/cos(x)) / (t - x) h->0 = lim (sen(t)cos(x) - sen(x)cos(t)) / (t - x)cos(t)cos(x) h->0 = lim 1 / cos^2(x) h->0 = sec^2(x) 4) f(x) = cossec(x) f'(x) = lim (cossec(x+h) - cossec(x)) / h h->0 = lim (1/sen(x+h) - 1/sen(x)) / h h->0 = lim (sen(x) - [sen(x)cos(h) + sen(h)cos(x)]) / [sen(x+h)sen(x)] * 1/h h->0 = lim -cos(x) sen(h) / sen(x)sen(x) * 1/h h->0 = lim 1 / sen(x) cos(x) / sen(x) h->0 = -cossec(x) cotg(x) 5) f(x) = sec(x) f'(x) = lim (sec(x+h) - sec(x)) / h h->0 = lim (1/cos(x+h) - 1/cos(x)) / h h->0 = lim (cos(x) - [cos(x)cos(h) - sen(x)sen(h)]) / [cos(x+h)cos(x)] * 1/h h->0 = lim sen(x) sen(h) / cos(x)cos(x) * 1/h h->0 = lim 1 / cos^2(x) * sen(x)/cos(x) h->0 = sec(x) tg(x) 6) f(x) = cotg(x) f'(x) = lim (cotg(x+h) - cotg(x)) / h h->0 = lim (cos(t) - cos(x)) / sen(t)sen(x) / (t - x) t->x = lim sen(x)cos(t) - sen(t)cos(x) / sen(t)sen(x) * 1 / (t - x) = lim sen(x-t) / sen(x)sen(x) = lim -1/sen^2(x) = -cossec^2(x) 7) f(x) = arcsen(x) arcsen'(x) = 1 / sen'(arcsen(x)) => arcsen'(x) = 1 / cos(arcsen(x)) Pela relação fundamental da trigonometria, temos: [sen(arcsen(x))]^2 + [cos(arcsen(x))]^2 = 1 => [cos(arcsen(x))]^2 = 1 - x^2 Elevando ao quadrado (I): [arcsen'(x)]^2 = 1 / [cos(arcsen(x))]^2 => arcsen'(x) = 1 / sqrt(1-x^2), x ∈ [-π/2, π/2] 4) arcosec'(x) = \frac{1}{cosec'(arcosec(x))} => arcosec'(x) = \frac{1}{-cosec(arcosec(x)) \cdot cotg(arcosec(x))}\^x Chomando arcosec(x) = \theta, temos: scp^2 \theta + cos^2 \theta = 1 \div(sc^2 \theta) => 1 + \frac{cos^2 \theta}{sin^2 \theta} = \frac{1}{sin^2 \theta} => cotg^2 \theta = cosec ^2 \theta - 1 mas cotg^2(arcosec(x)) = cosec^2(arcosec(x)) - 1 ) => cotg(arcosec(x)) = \sqrt{x^2 - 1}, x \in \left[ \frac{-\Pi}{2}, \frac{\Pi}{2} \right] <math xmlns="http://www.w3.org/1998/Math/MathML"><invisible_times /></math> => arcosec(x) = \frac{-cosec(arcosec(x)) \cdot cotg(arcosec(x))}{x} \therefore(1) => arcosec(x) = \frac{1}{{-x\sqrt{x^2-1}}}, x \in \left[ \frac{-\Pi}{2}, \frac{\Pi}{2} \right] 5) arcsec'(x) = \frac{1}{sec'(arcsec(x))} => arcsec'(x) = {\frac{1}{sec(arcsec(x)) \cdot tg(arcsec(x))}}^x (I) Chomando arcsec(x) = \theta, temos: sin^2 \theta + cos^2 \theta = 1 \div(cos^2 \theta) => \frac{sin^2 \theta}{cos^2 \theta} + 1 = \frac{1}{cos^2 \theta} => sec^2 \theta = sec\theta - 1 mas tg^2(arcsec(x)) = sec^2(arcsec(x)) - 1 \div sec^2(arcsec(x)) - 1 => tg(arcsec(x)) = \sqrt{x^2 - 1}, x \in \left[ \frac{-\Pi}{2}, \frac{\Pi}{2} \right] <math xmlns="http://www.w3.org/1998/Math/MathML"><invisible_times /></math> (1) => arcsec(x) = \frac{1}{{x\sqrt{x^2-1}}}, x \in \left[ \frac{-\Pi}{2}, \frac{\Pi}{2} \right] 6) arccotg(x) = \frac{1}{cotg(arcocotg(x))} => arccotg'(x) = \frac{1}{-cosec^2(arcocotg(x))} (I) Chamando arccotg(x) = \theta, temos: sc\theta + cos^2 \theta = 1 \div(sc^2 \theta) => \frac{sin^2 \theta}{sec^2 \theta} + \frac{1}{sec^2 \theta} \frac{1}{sec^2 \theta} => cosec^2\theta = 1 + cotg^2 \theta => cosec^2(arccotg(x)) = 1 + cotg(arccotg(x)) => cotg(arccotg(x)) = cosec^2(arccotg(x)) = 1+x^2 math </math\n\therefore (I) => arccotg(x) = \frac{-1}{1+x^2}, x \in \left[ \frac{-\Pi}{2}, \frac{\Pi}{2} \right]