derivative of 4/(2x+1)

Lösung

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

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

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende Exponentenregel an:

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

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

t an g e n t f (x) = 2 x^{2} + x - 1, at x = 0

\frac{d}{d x} (lo g_{10} (\frac{x}{x ^{2} + 1}))

\frac{\partial}{\partial x} (15 - 2 x^{2} - y^{2})

d er i v a t i v e (3 + x^{4})^{\frac{4}{5}}

x \to 0 + lim (\frac{10}{x ^{2}})