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

Lösung

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

- \frac{5 y}{x ^{6}}

Schritte zur Lösung

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

Behandle

y

als Konstante

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{- 2}, g = \frac{1}{x ^{3}}

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

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

\frac{\partial}{\partial x} (\frac{1}{x ^{3}}) = - \frac{3}{x ^{4}}

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

Vereinfache

y ((- \frac{2}{x ^{3}}) \frac{1}{x ^{3}} + (- \frac{3}{x ^{4}}) x^{- 2}) : - \frac{5 y}{x ^{6}}

= - \frac{5 y}{x ^{6}}

y y^{^{'}} + x y^{2} = x y^{4}

\int sec^{8} (x) tan (x) d x

\int - 9 e^{- x} d x

y^{^{''}} - 2 y^{^{'}} + y = e^{3 x}

\frac{d}{d x} (r^{2} - 19 x^{2})