(\partial)/(\partial x)((3+x^2-3y^2)^{1/2})

Lösung

\frac{\partial}{\partial x} ((3 + x^{2} - 3 y^{2})^{\frac{1}{2}})

\frac{x}{( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}}

Schritte zur Lösung

\frac{\partial}{\partial x} ((3 + x^{2} - 3 y^{2})^{\frac{1}{2}})

Behandle

y

als Konstante

Wende die Kettenregel an:

\frac{1}{2 ( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}} \frac{\partial}{\partial x} (3 + x^{2} - 3 y^{2})

= \frac{1}{2 ( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}} \frac{\partial}{\partial x} (3 + x^{2} - 3 y^{2})

\frac{\partial}{\partial x} (3 + x^{2} - 3 y^{2}) = 2 x

= \frac{1}{2 ( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}} \cdot 2 x

Vereinfache

\frac{1}{2 ( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}} \cdot 2 x : \frac{x}{( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}}

= \frac{x}{( 3 + x ^{2} - 3 y ^{2} ) ^{\frac{1}{2}}}

t a y l or x^{2} e^{3 x}

y^{^{''}} - 6 y^{^{'}} - 7 y = 0, y (0) = b, y^{^{'}} (0) = 8

\int ln (x) x^{'} d x

\frac{d}{d x} (3 e^{x^{2} + x})

y y^{^{'}} = \frac{7}{x}