• Matéria: Matemática
  • Autor: nathanyoliveira1
  • Perguntado 8 anos atrás

Alguém me ajuda com a resolução dessa derivada de y=4x^3/2 - 5x^1/2

Respostas

respondido por: andresccp
1
\text{soma de fracoes} \to \frac{A}{B} \pm \frac{C}{D}= \frac{AD \pm CB}{BD} \\\\ \text{derivada da potencia }\to (U^N)' = N*U^{N-1}*U'

  \\ y=4x^{ \frac{3}{2} }-5x^{ \frac{1}{2} }\\\\y'=4* \frac{3}{2} x^{( \frac{3}{2}-1) }  - 5* \frac{1}{2} x^{( \frac{1}{2} -1)}\\\\ y'= \frac{12}{2} x^{ \frac{1}{2} }  -  \frac{5}{2} x^{-\frac{1}{2}}\\\\y' = \frac{12x^{ \frac{1}{2} }}{2} - \frac{5}{2}* \frac{1}{x^{ \frac{1}{2} }}  \\\\y'= \frac{12 \sqrt{x} }{2} - \frac{5}{2 \sqrt{x} } \\\\y'= \frac{12\sqrt{x}*\sqrt{x}-5}{2\sqrt{x}} \\\\y'= \frac{12x-5}{2\sqrt{x}}

nathanyoliveira1: Obrigada :)
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