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sf33
@sf33
October 2020
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me ajudemmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm.
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jacksonmayson
1) f'= 3x^2 y^3Cos(x^3y^3). f''=6xy^3.cos(x^3y^3)+3x^2 y^3 -3x^2y^3sen(x^3y^3)
f''= 6xy^3cos(x^3 y^3)- 9x^4 y^6 sen (x^3 y^3)
alternativa A.
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sf33
muitissimo obrigado.
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Sf33
April 2020 | 0 Respostas
Por favor me ajudem com estes calculo urgenteeeeeeee.
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Sf33
April 2020 | 0 Respostas
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Lista de comentários
f''= 6xy^3cos(x^3 y^3)- 9x^4 y^6 sen (x^3 y^3)
alternativa A.