Question

derivadas Ayuda Plis

Answer 1

ula primera
y′= (x3)′= 
=3·(x)′·x2=3·1·x2= 
y'=3·x^2
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la segunda
y′= (5)′= 
y'=0
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la tercera
y′= (7·x2)′= 
=7·(x2)′=7·2·(x)′·x1= 
=14(x)′·x=14·1·x= 
y'=14x
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la cuarta
y′= (a·x4−(b·x2))′= 
=(a·x4)′−(b·x2)′= 
=a·(x4)′−(b·(x2)′)=a·4·(x)′·x3−(b·2·(x)′·x1)=
=4·a·(x)′·x3−(2b(x)′x)=4·a·1·x3−(2·b·1·x)=
y'=4ax^3−(2bx)
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la quinta
y′=(x^(4/3)+5)′=
=(5+3√x4)′= 
=(5)′+(3√x4)′=0+43(x)′·3√x= 
=43(x)′·3√x=43·1·3√x= 
y'=43·3√x
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la sexta
y′= ((x^2−3)^5)′= 
=5·(x^2−3)′·(x^2−3)4= 
=5·((x^2)′−(3)′)·(x^2−3)4=5·(2(x)′·x1−0)·(x^2−3)4=
=10·(x^2−3)4(x)′·x=10·(x^2−3)4·1·x= 
y'=10(x^2−3)4x
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la septima
y′=(3x^4−(2x^2)+8)′=
=(8+3x^4−(2x^2))′= 
=(8)′+(3x^4−(2x^2))′=0+(3x^4)′−(2x^2)′=
=(3x^4)′−(2x^2)′= 
=3·(x^4)′−(2·(x^2)′)=3·4·(x)′·x^3−(2·2·(x)′·x^1)=
=12·(x)′·x^3−(4·(x)′·x)=12·1·x3−(4·1·x)=
y'=12x^3−(4x)
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la octava
y′=(at^2−(5bt^3))′=
=(at^2−(5·t^3·b))′= 
=(at^2)′−(5·t^3·b)′= 
=a(t^2)′−(5(t^3·b)′)= 
=a·(2)′·Ln(t)·t^2−(5·b·(t^3)′)=a·0·Ln(t)·t^2−(5·b·(3)′·Ln(t)·t^3)=
=−5b·(3)′·Ln(t)·t^3=−5b·0·Ln(t)·t^3=
y'=0
asi se asen las otras saludos