Question

encontre o jacobiano de f(x,y)=(x^3;x+y)

Answer 1

O jacobiano de uma função $f:\mathbb{R}^{2}\to\mathbb{R}^{2}$ é definido por

$J_{f}(x,y)=\det\left[\begin{array}{cc}\dfrac{\partial f_{1}}{\partial x}&\dfrac{\partial f_{1}}{\partial y}\\\\\dfrac{\partial f_{2}}{\partial x}&\dfrac{\partial f_{2}}{\partial y}\end{array}\right]$

Onde $f_{1},\,f_{2}$ são as componentes de $f$. Ou seja, $f(x,y)=\big(f_{1}(x,y),\,f_{2}(x,y)\big)$
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Temos $f(x,y)=\big(x^{3},\,x+y\big)$, logo

$\bullet\,\,f_{1}(x,y)=x^{3}~~\Longrightarrow~~\dfrac{\partial f_{1}}{\partial x}(x,y)=3x^{2}~~~~e~~~~\dfrac{\partial f_{2}}{\partial y}(x,y)=0\\\\\\\bullet\,\,f_{2}(x,y)=x+y~~\Longrightarrow~~\dfrac{\partial f_{1}}{\partial x}(x,y)=1~~~e~~~\dfrac{\partial f_{2}}{\partial y}(x,y)=1$

Daí, o jacobiano de $f$ será o determinante:

$J_{f}(x,y)=\det\left[\begin{array}{cc}3x^{2}&0\\1&1\end{array}\right]\\\\\\J_{f}(x,y)=3x^{2}\times1-0\times1\\\\J_{f}(x,y)=3x^{2}-0\\\\\boxed{\boxed{J_{f}(x,y)=3x^{2}}}$