derivative of (3x/(sqrt(x^2+2)))

Lösung

\frac{d}{d x} (\frac{3 x}{x ^{2} + 2})

\frac{6}{( x ^{2} + 2 ) x ^{2} + 2}

Schritte zur Lösung

\frac{d}{d x} (\frac{3 x}{x ^{2} + 2})

Entferne die Konstante:

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

= 3 \frac{d}{d x} (\frac{x}{x ^{2} + 2})

Wende die Quotientenregel an:

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

= 3 \cdot \frac{\frac{d x}{d x} x ^{2} + 2 - \frac{d}{d x} ( x ^{2} + 2 ) x}{( x ^{2} + 2 ) ^{2}}

\frac{d x}{d x} = 1

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

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

Vereinfache

3 \cdot \frac{1 \cdot x ^{2} + 2 - \frac{x}{x ^{2} + 2} x}{( x ^{2} + 2 ) ^{2}} : \frac{6}{( x ^{2} + 2 ) x ^{2} + 2}

= \frac{6}{( x ^{2} + 2 ) x ^{2} + 2}

\frac{d}{d x} (x cos (6 x))

d er i v a t i v e y = \frac{2 ( 2 x ^{5} - 9 )}{x ^{4}}

\frac{d y}{d t} = y^{3}

y^{^{''}} + 2 y^{^{'}} + 2 y = 5 cos (3 t) - 20 sin (3 t)

t an g e n t f (x) = - x^{2} + 25, at x = 3