derivative (x+2)^{2^'}

Lösung

ableitung von

(x + 2)^{2^{^{'}}}

2^{'} (x + 2)^{2^{'} - 1}

Schritte zur Lösung

\frac{d}{d x} ((x + 2)^{2^{^{'}}})

Wende Exponentenregel an:

a^{b} = e^{b l n (a)}

(x + 2)^{2^{^{'}}} = e^{2^{^{'}} l n (x + 2)}

= \frac{d}{d x} (e^{2^{^{^{'}}} l n (x + 2)})

Wende die Kettenregel an:

e^{2^{^{'}} l n (x + 2)} \frac{d}{d x} (2^{^{'}} ln (x + 2))

= e^{2^{^{'}} l n (x + 2)} \frac{d}{d x} (2^{^{'}} ln (x + 2))

\frac{d}{d x} (2^{^{'}} ln (x + 2)) = \frac{2 ^{^{'}}}{x + 2}

= e^{2^{^{^{'}}} l n (x + 2)} \frac{2 ^{^{^{'}}}}{x + 2}

Vereinfache

e^{2^{^{'}} l n (x + 2)} \frac{2 ^{^{'}}}{x + 2} : 2^{^{'}} (x + 2)^{2^{^{'}} - 1}

= 2^{^{'}} (x + 2)^{2^{^{'}} - 1}

\int 9 e^{2 x} d x

\int_{1}^{11 x^{15}} \frac{1}{t} d t

\frac{d}{d x} (\frac{2 x ^{5}}{5 + 3 x ^{4} + 6 x ^{3} + 2 x + 4})

z \to 0 lim ((2 z - 8) \frac{1}{3})

\frac{d}{d x} (\frac{2 x}{3 - tan ( x )})