derivative of (e^{4x}/(ln(6x)))

Lösung

\frac{d}{d x} (\frac{e ^{4 x}}{ln ( 6 x )})

\frac{4 e ^{4 x} x ln ( 6 x ) - e ^{4 x}}{x ln ^{2} ( 6 x )}

Schritte zur Lösung

\frac{d}{d x} (\frac{e ^{4 x}}{ln ( 6 x )})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( e ^{4 x} ) ln ( 6 x ) - \frac{d}{d x} ( ln ( 6 x ) ) e ^{4 x}}{( ln ( 6 x ) ) ^{2}}

\frac{d}{d x} (e^{4 x}) = e^{4 x} \cdot 4

\frac{d}{d x} (ln (6 x)) = \frac{1}{x}

= \frac{e ^{4 x} \cdot 4 ln ( 6 x ) - \frac{1}{x} e ^{4 x}}{( ln ( 6 x ) ) ^{2}}

Vereinfache

\frac{e ^{4 x} \cdot 4 ln ( 6 x ) - \frac{1}{x} e ^{4 x}}{ln ^{2} ( 6 x )} : \frac{4 e ^{4 x} x ln ( 6 x ) - e ^{4 x}}{x ln ^{2} ( 6 x )}

= \frac{4 e ^{4 x} x ln ( 6 x ) - e ^{4 x}}{x ln ^{2} ( 6 x )}

\int \frac{2 x}{e ^{x}} d x

\frac{d}{d x} (36 x^{2})

\int 2 x (3 (x + 3)) d x

5 y^{^{''}} + 60 y = 0

\int_{0}^{\frac{1}{2}} \frac{4}{1 + 4 t ^{2}} d t