(\partial)/(\partial x)((4x)/(sqrt(x^2+2)))

Lösung

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

\frac{8}{( x ^{2} + 2 ) x ^{2} + 2}

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende die Quotientenregel an:

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

= 4 \cdot \frac{\frac{\partial}{\partial x} ( x ) x ^{2} + 2 - \frac{\partial}{\partial x} ( x ^{2} + 2 ) x}{( x ^{2} + 2 ) ^{2}}

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

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

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

Vereinfache

4 \cdot \frac{1 \cdot x ^{2} + 2 - \frac{x}{x ^{2} + 2} x}{( x ^{2} + 2 ) ^{2}} : \frac{8}{( x ^{2} + 2 ) x ^{2} + 2}

= \frac{8}{( x ^{2} + 2 ) x ^{2} + 2}

f (x) = 5 x + 1

\int sec^{4} (3 x) tan^{3} (3 x) d x

\frac{d}{d x} (6 e^{- x})

a re a y = 5 x, y = \frac{1}{7} x, y = 50 - x^{2}

\frac{d}{d θ} (sin (θ) + tan (θ))