derivative of log_{2}(x+1)

Lösung

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

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

Schritte zur Lösung

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

Wende die log Regel an:

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

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

Entferne die Konstante:

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

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

(x^{4})^{^{'}}

\frac{\partial}{\partial x} (\frac{1}{2} y arctan (\frac{x}{y}))

\int \frac{x ^{2} - 3 x - 7}{( 2 x + 3 ) ( x + 1 )} d x

\int_{1}^{e^{3}} \frac{1}{x ( 1 + ln ( x ) )} d x

y^{^{''}} + 25 y = 4 csc^{2} (5 t)