derivative of (40/(0.01x^2+1))

Lösung

\frac{d}{d x} (\frac{40}{0.01 x ^{2} + 1})

- \frac{0.8 x}{( 0.01 x ^{2} + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{40}{0.01 x ^{2} + 1})

Entferne die Konstante:

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

= 40 \frac{d}{d x} (\frac{1}{0.01 x ^{2} + 1})

Wende Exponentenregel an:

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

= 40 \frac{d}{d x} ((0.01 x^{2} + 1)^{- 1})

Wende die Kettenregel an:

- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \frac{d}{d x} (0.01 x^{2} + 1)

= - \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \frac{d}{d x} (0.01 x^{2} + 1)

\frac{d}{d x} (0.01 x^{2} + 1) = 0.02 x

= 40 (- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \cdot 0.02 x)

Vereinfache

40 (- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \cdot 0.02 x) : - \frac{0.8 x}{( 0.01 x ^{2} + 1 ) ^{2}}

= - \frac{0.8 x}{( 0.01 x ^{2} + 1 ) ^{2}}

\frac{d}{d x} (e^{n x^{2}})

x^{^{'}} = x^{2} + x

\frac{d}{d x} (lo g_{7} (e^{- x} cos (π x)))

t an g e n t y = \frac{1}{x + 1}, (1, \frac{1}{2})

l a pl a ce t r an s f or m (1 + t)^{3}