(\partial ^2)/(\partial x\partial y)(ln(3x^2-y^3-3))

Lösung

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

\frac{18 x y ^{2}}{( 3 x ^{2} - y ^{3} - 3 ) ^{2}}

Schritte zur Lösung

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

\frac{\partial}{\partial x} (ln (3 x^{2} - y^{3} - 3)) = \frac{6 x}{3 x ^{2} - y ^{3} - 3}

= \frac{\partial}{\partial y} (\frac{6 x}{3 x ^{2} - y ^{3} - 3})

\frac{\partial}{\partial y} (\frac{6 x}{3 x ^{2} - y ^{3} - 3}) = \frac{18 x y ^{2}}{( 3 x ^{2} - y ^{3} - 3 ) ^{2}}

= \frac{18 x y ^{2}}{( 3 x ^{2} - y ^{3} - 3 ) ^{2}}

\frac{d}{d x} (9 e^{- x} + e^{4 x})

\int (\frac{x ^{3}}{x ^{2} + 1}) d x

\frac{\partial}{\partial z} (3 x yz)

\frac{d}{d x} ((2 x^{2} - 1)^{3} \cdot (1 - x^{2})^{- 3})

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