derivative of 1/(2x+sin^3(x))

Lösung

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

- \frac{2 + 3 sin ^{2} ( x ) cos ( x )}{( 2 x + sin ^{3} ( x ) ) ^{2}}

Schritte zur Lösung

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

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

= - \frac{2 + 3 sin ^{2} ( x ) cos ( x )}{( 2 x + sin ^{3} ( x ) ) ^{2}}

\frac{d}{d x} ((2 x^{2} - 6 x + 1)^{- 8})

\int \frac{1}{2 x - 1} d x

y^{^{''}} + 16 y = 8 tan (4 x)

\frac{2 y ^{^{'}}}{y} = x

\int \frac{2 x ^{3} - 4 x ^{2} - x - 3}{x ^{2} - 2 x - 3} d x