Considere a função multivariada
f left parenthesis x comma y right parenthesis equals e to the power of 2 x end exponent sin y left parenthesis 4 x y squared plus 2 right parenthesis
Denote a derivada parcial de f left parenthesis x comma y right parenthesis em relação a x como f subscript x, e a derivada parcial de f left parenthesis x comma y right parenthesis em relação a y como f subscript y. Calcule f subscript x e f subscript y.
Escolha a alternativa correta.
Escolha uma:
a.
f subscript x equals e to the power of 2 x end exponent left square bracket sin y left parenthesis 4 x y squared plus 2 x right parenthesis plus sin y 8 x y right square bracket, e f subscript y equals e to the power of 2 x end exponent sin y 2 left square bracket 2 y squared left parenthesis 2 x plus 1 right parenthesis plus x plus 1 right square bracket.
b.
f subscript x equals sin y 2 left square bracket 2 y squared left parenthesis 2 x plus 1 right parenthesis plus x plus 1 right square bracket, e f subscript y equals cos y left parenthesis 4 x y squared plus 2 x right parenthesis plus sin y 8 x y.
c.
f subscript x equals e to the power of 2 x end exponent sin y 2 left square bracket 2 left parenthesis 2 x plus 1 right parenthesis plus x plus 1 right square bracket, e f subscript y equals e to the power of 2 x end exponent left square bracket cos y 2 x plus sin y 8 x y right square bracket.
d.
f subscript x equals e to the power of 2 x end exponent sin y 2 left square bracket 2 y squared left parenthesis 2 x plus 1 right parenthesis plus x plus 1 right square bracket, e f subscript y equals e to the power of 2 x end exponent left square bracket cos y left parenthesis 4 x y squared plus 2 x right parenthesis plus sin y 8 x y right square bracket.
e.
f subscript x equals e to the power of 2 x end exponent cos y left square bracket 2 y squared left parenthesis 2 x plus 1 right parenthesis plus x plus 1 right square bracket, e f subscript y equals e to the power of 2 x end exponent left square bracket sin y left parenthesis x y squared plus 2 x right parenthesis plus sin y 8 x y right square bracket.
Respostas
respondido por:
9
A resposta correta é:
fx = e^2x siny2 [2y^2(2x+1) + x + 1] e fy = e^2x [cosy(4xy^2 + 2x) + siny8xy]
fx = e^2x siny2 [2y^2(2x+1) + x + 1] e fy = e^2x [cosy(4xy^2 + 2x) + siny8xy]
winiciussouza2p92s9l:
CORRETO! CORRIGIDO PELO AVA
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