((100)/(log_{10)(2x+1)})^'

Lösung

(\frac{100}{lo g _{10} ( 2 x + 1 )})^{^{'}}

- \frac{200}{ln ( 10 ) lo g _{10} ( 2 x + 1 ) ^{2} ( 2 x + 1 )}

Schritte zur Lösung

(\frac{100}{lo g _{10} ( 2 x + 1 )})^{'}

Entferne die Konstante:

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

= 100 (\frac{1}{lo g _{10} ( 2 x + 1 )})^{^{'}}

Wende Exponentenregel an:

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

= 100 (lo g_{10} (2 x + 1)^{- 1})^{^{'}}

Wende die Kettenregel an:

- \frac{1}{( lo g _{10} ( 2 x + 1 ) ) ^{2}} (lo g_{10} (2 x + 1))^{'}

= - \frac{1}{( lo g _{10} ( 2 x + 1 ) ) ^{2}} (lo g_{10} (2 x + 1))^{'}

(lo g_{10} (2 x + 1))^{'} = \frac{2}{ln ( 10 ) ( 2 x + 1 )}

= 100 (- \frac{1}{( lo g _{10} ( 2 x + 1 ) ) ^{2}} \cdot \frac{2}{ln ( 10 ) ( 2 x + 1 )})

Vereinfache

100 (- \frac{1}{lo g _{10} ( 2 x + 1 ) ^{2}} \cdot \frac{2}{ln ( 10 ) ( 2 x + 1 )}) : - \frac{200}{ln ( 10 ) lo g _{10} ( 2 x + 1 ) ^{2} ( 2 x + 1 )}

= - \frac{200}{ln ( 10 ) lo g _{10} ( 2 x + 1 ) ^{2} ( 2 x + 1 )}

f (x) = x^{3} e^{2 x}

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

y^{^{''}} - 5 y^{^{'}} - 6 y = 0, y (0) = 5, y^{^{'}} (0) = 12

\int \frac{1}{10 x ^{\frac{1}{2}} - x} d x

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