derivative of log_{2}(\sqrt[3]{2x+1})

Lösung

\frac{d}{d x} (lo g_{2} (3 2 x + 1))

\frac{2}{3 ln ( 2 ) 3 2 x + 1 ( 2 x + 1 ) ^{\frac{2}{3}}}

Schritte zur Lösung

\frac{d}{d x} (lo g_{2} (3 2 x + 1))

Wende die log Regel an:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( 3 2 x + 1 )}{ln ( 2 )})

Entferne die Konstante:

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

= \frac{1}{ln ( 2 )} \frac{d}{d x} (ln (3 2 x + 1))

Wende die Kettenregel an:

\frac{1}{3 2 x + 1} \frac{d}{d x} (3 2 x + 1)

= \frac{1}{3 2 x + 1} \frac{d}{d x} (3 2 x + 1)

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

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

Vereinfache

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

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

\frac{d}{d x} (f (4 x^{5})) = 8 x^{4}

d er i v a t i v e cos (6 t)

d er i v a t i v e \frac{( 2 x + 7 )}{x}

3 y^{^{''}} + 3 y^{^{'}} - y = 0

t an g e n t x^{3} y^{2} = x y^{3} + 6, (2, 1)