(\partial)/(\partial x)((2F(V,x,P))/(V^2xP))

Lösung

\frac{\partial}{\partial x} (\frac{2 F ( V , x , P )}{V ^{2} x P})

\frac{2}{V ^{2} P} \cdot \frac{\frac{\partial}{\partial x} ( F ( V , x , P ) ) x - 1 \cdot F ( V , x , P )}{x ^{2}}

Schritte zur Lösung

\frac{\partial}{\partial x} (\frac{2 F ( V , x , P )}{V ^{2} x P})

Behandle

V, P

als Konstanten

Entferne die Konstante:

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

= \frac{2}{V ^{2} P} \frac{\partial}{\partial x} (\frac{F ( V , x , P )}{x})

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{2}{V ^{2} P} \cdot \frac{\frac{\partial}{\partial x} ( F ( V , x , P ) ) x - \frac{\partial}{\partial x} ( x ) F ( V , x , P )}{x ^{2}}

\frac{\partial}{\partial x} (x) = 1

= \frac{2}{V ^{2} P} \cdot \frac{\frac{\partial}{\partial x} ( F ( V , x , P ) ) x - 1 \cdot F ( V , x , P )}{x ^{2}}

\frac{d ^{3}}{d x ^{3}} (\frac{5}{x})

d er i v a t i v e y = (3 x^{6} - 2 x^{4})^{4}

\frac{\partial}{\partial y} (e^{x y z^{9}} y z^{9})

\int 2 e^{4} d x

25 \frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} + y = 0