Ensino Superior ⇒ Derivadas parciais

[Natan]

Seja [tex3]w=\frac{e^{x+y}}{e^x+e^y}.[/tex3] Mostre que [tex3]\frac{{\partial}w}{{\partial}x}+\frac{{\partial}w}{{\partial}y}=w[/tex3]

[Rafa2604]

Seja [tex3]w=\frac{e^{x+y}}{e^x+e^y}[/tex3], então temos que:

[tex3]\frac{\partial w}{\partial x} = \frac{\partial}{\partial x}\left(\frac{e^{x+y}}{e^x+e^y}\right) = \text{regra do quociente} = \frac{e^{x+y}.(e^{x}+e^{y}) - e^{x+y}(e^{x}) }{(e^{x}+e^{y})^{2}} = \frac{e^{2x+y}+ e^{x+2y} - e^{2x+y} }{(e^{x}+e^{y})^{2}} = \frac{e^{x+2y}}{(e^{x}+e^{y})^{2}}[/tex3]

[tex3]\frac{\partial w}{\partial y} = \frac{\partial}{\partial y}\left(\frac{e^{x+y}}{e^x+e^y}\right) = \text{regra do quociente} = \frac{e^{x+y}.(e^{x}+e^{y}) - e^{x+y}(e^{y}) }{(e^{x}+e^{y})^{2}} = \frac{e^{2x+y}+ e^{x+2y} - e^{x+2y} }{(e^{x}+e^{y})^{2}} = \frac{e^{2x+y}}{(e^{x}+e^{y})^{2}}[/tex3]

Logo, temos que: [tex3]\;\frac{{\partial}w}{{\partial}x}+\frac{{\partial}w}{{\partial}y}= \frac{e^{x+2y}}{(e^{x}+e^{y})^{2}} + \frac{e^{2x+y}}{(e^{x}+e^{y})^{2}} = \frac{e^{x+2y}+e^{2x+y}}{(e^{x}+e^{y})^{2}} = \frac{(e^{x+y}) (e^{y}+e^{x})}{(e^{x}+e^{y})^{2}} = \frac{e^{x+y}}{e^{x}+e^{y}} = w[/tex3]