como resolver derivadas algebraicas de f(x)=x½+x⅓
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Respuesta:
F(X) = X^(1/2)+X^(1/3)
F ' (X) = d/dx [ X^(1/2)+X^(1/3) ]
F ' (X) = d/dx [ X^(1/2) ] + d/dx [ X^(1/3) ]
F ' (X) = (1/2)X^((1/2)-1)+(1/3)X^((1/3)-1)
F ' (X) = (1/2)X^(-1/2) + (1/3)X^(-2/3)
F ' (X) = 1/(2X^(1/2))+1/(3X^(2/3))
F ' (X) = 1/(2√(x))+1/(3∛(x)^2)
F ' (X) = (1/(2√(x))+(1/(3∛(x)^2))
R// " F ' (X) = (1/(2√(x))+(1/(3∛(x)^2)) " es la derivada de la función " F(X) = X^(1/2)+X^(1/3) " .
Explicación paso a paso:
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