derivative (60)/(0.01x^2+1)

Lösung

ableitung von

\frac{60}{0.01 x ^{2} + 1}

- \frac{1.2 x}{( 0.01 x ^{2} + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{60}{0.01 x ^{2} + 1})

Entferne die Konstante:

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

= 60 \frac{d}{d x} (\frac{1}{0.01 x ^{2} + 1})

Wende Exponentenregel an:

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

= 60 \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

= 60 (- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \cdot 0.02 x)

Vereinfache

60 (- \frac{1}{( 0.01 x ^{2} + 1 ) ^{2}} \cdot 0.02 x) : - \frac{1.2 x}{( 0.01 x ^{2} + 1 ) ^{2}}

= - \frac{1.2 x}{( 0.01 x ^{2} + 1 ) ^{2}}

x \to 2 lim (\frac{( 2 x ^{2} + 1 )}{( 3 x - 2 )})

d er i v a t i v e y = 1 6^{s i n (x^{2})}

s l o p e (2.3) (- 2.3)

\int \frac{x ^{5} - 9 x ^{3} - 9 x ^{2} - 9}{x ^{3} + 9 x} d x

d er i v a t i v e f (x) = tan (\frac{3 π}{4} x)