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Version vom 28. Februar 2022, 16:43 Uhr von Borishaase (Diskussion | Beiträge) (Leibnizsche Differentiationsregel)
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Leibnizsche Differentiationsregel

Für [math]\displaystyle{ f: {}^{(\omega)}\mathbb{K}^{n+1} \rightarrow {}^{(\omega)}\mathbb{K}, a, b: {}^{(\omega)}\mathbb{K}^{n} \rightarrow {}^{(\omega)}\mathbb{K}, \curvearrowright B x := {(s, {x}_{2}, ..., {x}_{n})}^{T} }[/math] und [math]\displaystyle{ s \in {}^{(\omega)}\mathbb{K} \setminus \{{x}_{1}\} }[/math] gilt bei Wahl von [math]\displaystyle{ \curvearrowright D a(x) = a(\curvearrowright B x) }[/math] und [math]\displaystyle{ \curvearrowright D b(x) = b(\curvearrowright B x) }[/math],

[math]\displaystyle{ \frac{\partial }{\partial {{x}_{1}}}\left( \int\limits_{a(x)}^{b(x)}{f(x,t)dDt} \right)=\int\limits_{a(x)}^{b(x)}{\frac{\partial f(x,t)}{\partial {{x}_{1}}}dDt}+\frac{\partial b(x)}{\partial {{x}_{1}}}f(\curvearrowright Bx,b(x))-\frac{\partial a(x)}{\partial {{x}_{1}}}f(\curvearrowright Bx,a(x)). }[/math]

Beweis:

[math]\displaystyle{ \begin{aligned}\frac{\partial }{\partial {{x}_{1}}}\left( \int\limits_{a(x)}^{b(x)}{f(x,t)dDt} \right) &={\left( \int\limits_{a(\curvearrowright Bx)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t)dDt}-\int\limits_{a(x)}^{b(x)}{f(x,t)dDt} \right)}/{\partial {{x}_{1}}}\;={\left( \int\limits_{a(x)}^{b(x)}{(f(\curvearrowright Bx,t)-f(x,t))dDt}+\int\limits_{b(x)}^{b(\curvearrowright Bx)}{f(\curvearrowright Bx,t)dDt}-\int\limits_{a(x)}^{a(\curvearrowright Bx)}{f(\curvearrowright Bx,t)dDt} \right)}/{\partial {{x}_{1}}}\; \\ &=\int\limits_{a(x)}^{b(x)}{\frac{\partial f(x,t)}{\partial {{x}_{1}}}dDt}+\frac{\partial b(x)}{\partial {{x}_{1}}}f(\curvearrowright Bx,b(x))-\frac{\partial a(x)}{\partial {{x}_{1}}}f(\curvearrowright Bx,a(x)).\square\end{aligned} }[/math]

Leseempfehlung

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