derivative 1/(3x^2+4x)

Lösung

ableitung von

\frac{1}{3 x ^{2} + 4 x}

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

Schritte zur Lösung

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

Wende Exponentenregel an:

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

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

\frac{d}{d x} ((4 x^{2} + 3 x + 2)^{5})

\int_{0}^{2} \frac{6 x}{( 1 + x ^{2} ) ^{2}} d x

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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = t^{- 2} e^{- 3 t}