derivative of log_{2x+1}(2x+3)

Lösung

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

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

Schritte zur Lösung

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

Wende die log Regel an:

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

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

\frac{\partial}{\partial x} (3 e^{x^{3} y^{3}})

t an g e n t 7 x + 4

\int 2 x (x^{2} + 2)^{4} d x

l a pl a ce t r an s f or m 7 sin (4 t + 8) + 3 cos (4 t - 2)

x \to 1 lim (1 - x - [x])