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La dérivée de f(x,y): $$f'(x,y)\ =\ (x^2+y^2)'$$ $$ f'_{x}\ = \left(\frac{\partial x^2+y^2}{\partial x} \right)_{y = constante}\ =\ \left(\frac{\partial x^2+Constante^2}{\partial x}\right)\ =\ 2.x $$ $$f'_{y}\ = \left(\frac{\partial x^2+y^2}{\partial y} \right)_{x = constante}\ =\ \left(\frac{\partial Constante^2+y^2}{\partial y}\right)\ =\ 2.y $$
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$\dfrac{\partial f}{\partial x} =3y$, $\dfrac{\partial f}{\partial y} =3x+2y$ et $\dfrac{\partial f}{\partial z} =-3z^2$.
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$\dfrac{\partial f}{\partial x} =\ e^{x}.\cos(y)$, $\dfrac{\partial f}{\partial y} =\ -e^{x}.\sin(y)$
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$\dfrac{\partial f}{\partial x} =\ 2x.\cos(xy)\ +\ \left(x^{2}\ +\ y^{2}\right).\left(-y.\sin(xy)\right)\ =\ 2x.\cos(xy)\ -\ y\left(x^{2}\ +\ y^{2}\right).\sin(xy)$ $\dfrac{\partial f}{\partial y} =\ 2y.\cos(xy)\ +\ \left(x^{2}\ +\ y^{2}\right).\left(-x.\sin(xy)\right)\ =\ 2y.\cos(xy)\ -\ x\left(x^{2}\ +\ y^{2}\right).\sin(xy)$
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$\dfrac{\partial f}{\partial x} =\ 2x.y^{2}.\frac{1}{2\sqrt{1\ +\ x^{2}.y^{2}}}\ =\ \frac{x.y^{2}}{\sqrt{1\ +\ x^{2}.y^{2}}}$ $\dfrac{\partial f}{\partial y} =\ 2y.x^{2}.\frac{1}{2\sqrt{1\ +\ x^{2}.y^{2}}}\ =\ \frac{x^{2}.y}{\sqrt{1\ +\ x^{2}.y^{2}}}$