Resposta
[tex3]-4xy\sen(x^2-y^2)^2[/tex3]
Como se deu a passagem para essa igualdade ? [tex3]=F(x^2-y^2)-F(1)-G(0)(x^2-y^2)+G(0)[/tex3]rcompany escreveu: ↑07 Set 2021, 01:44 [tex3]\begin{array}{rl}f(x,y)&=\displaystyle\int_1^{x^2-y^2}\left[\int_0^u\sin(t^2)dt\right]du\\[12pt]
&=\displaystyle\int_1^{x^2-y^2}(G(u)-G(0))du\quad G\text{ primitiva de }\!\sin(t^2)\\
&=F(x^2-y^2)-F(1)-G(0)(x^2-y^2)+G(0)\quad F\text{ primitiva de }G\\[24pt]
\dfrac{\partial f(x,y)}{\partial x}&=2xF^\prime(x^2-y^2)-2G(0)x\\
\dfrac{\partial^{2}f(x,y)}{\partial x\partial y}&=-4xyF^{\!\prime\prime}\!(x^2-y^2)=-4xy\sin\left(\left(x^2-y^2\right)^2\right)\\[12pt]
\end{array}
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