• Matéria: Matemática
  • Autor: julioram1911
  • Perguntado 4 anos atrás

ajudem a resolver por favor familia

Anexos:

Respostas

respondido por: Worgin
1

\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\\\ \lim_{h \to 0} \frac{cos(x+h)-cos(x)}{h}\\\\ \lim_{h \to 0} \frac{cos(x)cos(h)-sen(x)sen(h)-cos(x)}{h}\\\\ \lim_{h \to 0} \frac{cos(x)(cos(h)-1)-sen(x)sen(h)}{h}\\\\ \lim_{h \to 0} \frac{cos(x)(cos(h)-1)}{h}-\lim_{h \to 0}\frac{sen(x)sen(h)}{h}\\\\\lim_{h \to 0} cos(x).\lim_{h \to 0} \frac{(cos(h)-1)}{h}-\lim_{h \to 0} sen(x).\lim_{h \to 0}\frac{sen(h)}{h}\\\\\lim_{h \to 0} cos(x).0-\lim_{h \to 0} sen(x).1\\\\f'(x)=-sen(x)\\\\f'(\pi)=-sen(\pi)=0

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\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0} \\\\ \lim_{x \to x_0} \frac{(8x-2x^2)-(8x_0-2x_0^2)}{x-x_0} \\\\ \lim_{x \to \pi} \frac{(8x-2x^2)-(8\pi-2\pi^2)}{x-\pi} \\\\ \lim_{x \to \pi} \frac{8x-8\pi-2x^2+2\pi^2}{x-\pi} \\\\ \lim_{x \to \pi} \frac{(x-\pi)(8)-(x^2-\pi^2)(2)}{x-\pi} \\\\ \lim_{x \to \pi} \frac{(x-\pi)(8)-(x+\pi)(x-\pi)(2)}{x-\pi} \\\\ \lim_{x \to \pi} \frac{(x-\pi)(8-2(x+\pi))}{x-\pi} \\\\ \lim_{x \to \pi} 8-2(x+\pi)\\\\ \lim_{x \to \pi} 8-2x-2\pi\\\\ \lim_{x \to \pi} 8-4\pi

Só repetir o processo com -sen(x) e 8-4x pra saber o valor da derivada de segunda ordem.

respondido por: arochaaraujo1
0

Resposta:

Explicação passo-a-passo:

a) f(x) = cos (x)

f(x) = cos (π)

f'(x) = - sen (\pi)

f"(x) = - cos (\pi)

b) f(x) = 8x - 2x²

f'(x) = 8 - 4x

f"(x) = - 4

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