Question

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

Answer 1

Respuesta:

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

Explicación paso a paso:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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