(\partial)/(\partial x)(in(x+y)/(x-y))

Lösung

\frac{\partial}{\partial x} (in \frac{x + y}{x - y})

- \frac{2 in y}{( x - y ) ^{2}}

Schritte zur Lösung

\frac{\partial}{\partial x} (in \frac{x + y}{x - y})

Behandle

i, n, y

als Konstante

Entferne die Konstante:

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

= in \frac{\partial}{\partial x} (\frac{x + y}{x - y})

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

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

v^{'} = \frac{8 \cdot 9.8 + 3.1 v ^{2}}{8}

x \to 3 lim (\frac{x ^{2} + 1}{10})

y^{^{''}} - 5 y^{^{'}} = 8 t

y^{^{'}} = y, y (0) = 1