derivative of (ln(x^3+1)/(2^{x^4)-6})

Lösung

\frac{d}{d x} (\frac{ln ( x ^{3} + 1 )}{2 ^{x^{4}} - 6})

\frac{3 x ^{2} ( 2 ^{x^{4}} - 6 ) - ln ( 2 ) \cdot 2 ^{x^{4} + 2} x ^{3} ln ( x ^{3} + 1 ) ( x ^{3} + 1 )}{( x ^{3} + 1 ) ( 2 ^{x^{4}} - 6 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{ln ( x ^{3} + 1 )}{2 ^{x^{4}} - 6})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( ln ( x ^{3} + 1 ) ) ( 2 ^{x^{4}} - 6 ) - \frac{d}{d x} ( 2 ^{x^{4}} - 6 ) ln ( x ^{3} + 1 )}{( 2 ^{x^{4}} - 6 ) ^{2}}

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

\frac{d}{d x} (2^{x^{4}} - 6) = ln (2) \cdot 2^{x^{4} + 2} x^{3}

= \frac{\frac{3 x ^{2}}{x ^{3} + 1} ( 2 ^{x^{4}} - 6 ) - ln ( 2 ) \cdot 2 ^{x^{4} + 2} x ^{3} ln ( x ^{3} + 1 )}{( 2 ^{x^{4}} - 6 ) ^{2}}

Vereinfache

\frac{\frac{3 x ^{2}}{x ^{3} + 1} ( 2 ^{x^{4}} - 6 ) - ln ( 2 ) \cdot 2 ^{x^{4} + 2} x ^{3} ln ( x ^{3} + 1 )}{( 2 ^{x^{4}} - 6 ) ^{2}} : \frac{3 x ^{2} ( 2 ^{x^{4}} - 6 ) - ln ( 2 ) \cdot 2 ^{x^{4} + 2} x ^{3} ln ( x ^{3} + 1 ) ( x ^{3} + 1 )}{( x ^{3} + 1 ) ( 2 ^{x^{4}} - 6 ) ^{2}}

= \frac{3 x ^{2} ( 2 ^{x^{4}} - 6 ) - ln ( 2 ) \cdot 2 ^{x^{4} + 2} x ^{3} ln ( x ^{3} + 1 ) ( x ^{3} + 1 )}{( x ^{3} + 1 ) ( 2 ^{x^{4}} - 6 ) ^{2}}

\int \frac{1}{1 - u} d u

\int e^{7 x + 8} d x

d er i v a t i v e ln (tan^{2} (x))

a re a y = x, y = x^{2} - 2, y = 0

n = 0 \sum \infty e^{(2 n - 1)}