(4x^-3/4y^1/5)^-2

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Solution for (4x^-3/4y^1/5)^-2 equation: 


D( x )

x = 0

(((4*x^-3)/4)*y^1)/5 = 0

x = 0

x = 0

(((4*x^-3)/4)*y^1)/5 = 0

(((4*x^-3)/4)*y^1)/5 = 0

(((4*x^-3)/4)*y)/5 = 0

1/5*x^-3*y = 0 // : 1/5*y

x^-3 = 0

x należy do O

x in (-oo:0) U (0:+oo)

((((4*x^-3)/4)*y^1)/5)^-2 = 0

((((4*x^-3)/4)*y)/5)^-2 = 0

25*x^6*y^-2 = 0 // : 25*y^-2

x^6 = 0

x = 0

x in { 0} 

x belongs to the empty set

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