Matemática, perguntado por daniferreirra, 1 ano atrás

Ache todas as derivadas parciais de segunda ordem da função.
z= e^-xy^2

Soluções para a tarefa

Respondido por acidbutter
1
Em relação a X:
\displaystyle z(x,y)=e^{-xy^2}=\exp(-xy^2)\\\\ \frac{\partial }{\partial x}\exp(-xy^2)=\frac{\partial}{\partial u}\exp(u).\frac{\partial u}{\partial x}\implies (u(x,y)=-xy^2\\\\\implies \frac{\partial}{\partial x}-xy^2=-y^2\implies \frac{\partial z(x,y)}{\partial x}=\boxed{\exp(-xy^2).-y^2}\\\\\frac{\partial^2z(x,y)}{\partial x^2}=\frac{\partial}{\partial x}\exp(-xy^2).-y^2)\implies -y^2\frac{\partial}{\partial x}\exp(-xy^2)\implies
\displaystyle\frac{\partial^2z(x,y) }{\partial x^2}=-y^2.\exp(-xy^2).-y^2=\boxed{\exp(-xy^2)y^4}

Em relação a y:
\frac{\partial}{\partial y}\exp(-xy^2)=\frac{\partial}{\partial u}\exp(u)\cdot\frac{\partial u}{\partial y}\implies u(x,y)=-xy^2\implies \\\\\frac{\partial}{\partial y}(-xy^2)=-2xy\implies \frac{\partial}{\partial y}\exp(-xy^2)=\underline{\exp(-xy^2).-2xy}\implies \\\\ \frac{\partial^2 z(x,y)}{\partial y^2}=\frac{\partial}{\partial y}\exp(-xy^2).-2xy=-2x\frac{\partial}{\partial y}\exp(-xy^2)y\implies
\displaystyle \\\\-2x(\frac{\partial}{\partial y}\exp(-xy^2)\cdot y+\frac{\partial}{\partial y}y\cdot\exp(-xy^2))\implies \\\\\frac{\partial}{\partial y}\exp(-xy^2)=-2xy\cdot \exp(-xy^2)\\\\\frac{\partial}{\partial y}y=1\\\\\ \implies -2x(-2xy\cdot \exp(-xy^2)\cdot y+\exp(-xy^2))\implies \\-2x(\exp(-xy^2)-2\exp(-xy^2)\cdot xy^2)=\boxed{-2x(e^{-xy^2}-2e^{-xy^2}.xy^2)}ou seja:
\displaystyle \boxed{\frac{\partial^2}{\partial x^2}z(x,y)=\exp(-xy^2)y^4=e^{-xy^2}y^4}\\\boxed{\frac{\partial^2}{\partial y^2}z(x,y)=-2x(\exp(-xy^2)-2\exp(-xy^2)\cdot xy^2)}

daniferreirra: obrigado
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