(\partial)/(\partial y)(arctan(x^4y^3))

Lösung

\frac{\partial}{\partial y} (arctan (x^{4} y^{3}))

\frac{3 x ^{4} y ^{2}}{x ^{8} y ^{6} + 1}

Schritte zur Lösung

\frac{\partial}{\partial y} (arctan (x^{4} y^{3}))

Behandle

x

als Konstante

Wende die Kettenregel an:

\frac{1}{( x ^{4} y ^{3} ) ^{2} + 1} \frac{\partial}{\partial y} (x^{4} y^{3})

= \frac{1}{( x ^{4} y ^{3} ) ^{2} + 1} \frac{\partial}{\partial y} (x^{4} y^{3})

\frac{\partial}{\partial y} (x^{4} y^{3}) = 3 x^{4} y^{2}

= \frac{1}{( x ^{4} y ^{3} ) ^{2} + 1} \cdot 3 x^{4} y^{2}

Vereinfache

\frac{1}{( x ^{4} y ^{3} ) ^{2} + 1} \cdot 3 x^{4} y^{2} : \frac{3 x ^{4} y ^{2}}{x ^{8} y ^{6} + 1}

= \frac{3 x ^{4} y ^{2}}{x ^{8} y ^{6} + 1}

l a pl a ce t r an s f or m 5 (t - 1)

\frac{d}{d x} (\frac{A}{B + C e ^{x}})

\int e^{a x} d x

d er i v a t i v e f (x) = \frac{1 - x ^{2}}{2 x + 1}

2 t x^{^{''}} - x^{^{'}} + \frac{1}{( x ^{^{'}} )} = 0