• Asignatura: Matemáticas
  • Autor: marbanrubicelia03
  • hace 3 años

Si y=(x-2)^8/(3-x)^5 su derivada es

Respuestas

Respuesta dada por: S4NTA
0

Respuesta:

\frac{\left(x-2\right)^7\left(-3x+14\right)}{\left(3-x\right)^6}

Explicación paso a paso:

\left(\frac{\left(x-2\right)^8}{\left(3-x\right)^5}\right)'\:

\mathrm{Aplicar\:la\:regla\:del\:cociente}:\quad \left(\frac{f}{g}\right)^'=\frac{f\:'\cdot g-g'\cdot f}{g^2}

=\frac{\left(\left(x-2\right)^8\right)'\left(3-x\right)^5-\left(\left(3-x\right)^5\right)'\left(x-2\right)^8}{\left(\left(3-x\right)^5\right)^2}

\left(\left(x-2\right)^8\right)'\:

=\left(u^8\right)'\left(\left(x-2\right)\right)'\:

=8u^7\left(\left(x-2\right)\right)'\:

\mathrm{Sustituir\:en\:la\:ecuacion}\:u=\left(x-2\right)

=8\left(x-2\right)^7\left(x-2\right)'\:

\left(x-2\right)'\:=1

=8\left(x-2\right)^7

\left(\left(3-x\right)^5\right)'\:

=5\left(3-x\right)^4\left(3-x\right)'\:

=5\left(3-x\right)^4\left(-1\right)

=-5\left(3-x\right)^4

=\frac{8\left(x-2\right)^7\left(3-x\right)^5-\left(-5\left(3-x\right)^4\right)\left(x-2\right)^8}{\left(\left(3-x\right)^5\right)^2}

=\frac{8\left(x-2\right)^7\left(3-x\right)^5+5\left(3-x\right)^4\left(x-2\right)^8}{\left(\left(3-x\right)^5\right)^2}

=\frac{\left(3-x\right)^4\left(-2+x\right)^7\left(-3x+14\right)}{\left(\left(3-x\right)^5\right)^2}

=\frac{\left(-x+3\right)^4\left(x-2\right)^7\left(-3x+14\right)}{\left(-x+3\right)^{10}}

=\frac{\left(x-2\right)^7\left(-3x+14\right)}{\left(3-x\right)^6}

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