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

Lösung

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

- \frac{16 x}{( 4 x ^{2} + 1 ) ^{2}}

Schritte zur Lösung

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende Exponentenregel an:

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

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

Wende die Kettenregel an:

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

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

\frac{d}{d x} (4 x^{2} + 1) = 8 x

= 2 (- \frac{1}{( 4 x ^{2} + 1 ) ^{2}} \cdot 8 x)

Vereinfache

2 (- \frac{1}{( 4 x ^{2} + 1 ) ^{2}} \cdot 8 x) : - \frac{16 x}{( 4 x ^{2} + 1 ) ^{2}}

= - \frac{16 x}{( 4 x ^{2} + 1 ) ^{2}}

\frac{d x}{d t} = \frac{x ^{3} - x t ^{2}}{t ^{3}}, x (1) = 2

\int 3 x^{2} + 2 x + 4 d x

\frac{d}{d x} (2.1 + e^{2} - \frac{π ^{2}}{20})

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

y^{^{'}} = \frac{x ^{4} y ^{4} + x ^{4}}{y ^{3}}