Analysis Beispiele

x^{\sec (x)}

$x^{\sec (x)}$

Schritt 1

Wende die Logarithmengesetze an, um die Ableitung zu vereinfachen.

Tippen, um mehr Schritte zu sehen ...

Schritt 1.1

Schreibe

x^{\sec (x)}

$x^{\sec (x)}$ als

e^{\ln (x^{\sec (x)})}

$e^{\ln (x^{\sec (x)})}$ um.

\frac{d}{d x} [e^{\ln (x^{\sec (x)})}]

$\frac{d}{d x} [e^{\ln (x^{\sec (x)})}]$

Schritt 1.2

Zerlege

\ln (x^{\sec (x)})

$\ln (x^{\sec (x)})$ durch Herausziehen von

\sec (x)

$\sec (x)$ aus dem Logarithmus.

\frac{d}{d x} [e^{\sec (x) \ln (x)}]

$\frac{d}{d x} [e^{\sec (x) \ln (x)}]$

\frac{d}{d x} [e^{\sec (x) \ln (x)}]

$\frac{d}{d x} [e^{\sec (x) \ln (x)}]$

Schritt 2

Differenziere unter Anwendung der Kettenregel, die besagt, dass

\frac{d}{d x} [f (g (x))]

$\frac{d}{d x} [f (g (x))]$ ist

f' (g (x)) g' (x)

$f' (g (x)) g' (x)$ , mit

f (x) = e^{x}

$f (x) = e^{x}$ und

g (x) = \sec (x) \ln (x)

$g (x) = \sec (x) \ln (x)$ .

Tippen, um mehr Schritte zu sehen ...

Schritt 2.1

Um die Kettenregel anzuwenden, ersetze

u

$u$ durch

\sec (x) \ln (x)

$\sec (x) \ln (x)$ .

\frac{d}{d u} [e^{u}] \frac{d}{d x} [\sec (x) \ln (x)]

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [\sec (x) \ln (x)]$

Schritt 2.2

Differenziere unter Anwendung der Exponentialregel, die besagt, dass

\frac{d}{d u} [a^{u}]

$\frac{d}{d u} [a^{u}]$ gleich

a^{u} \ln (a)

$a^{u} \ln (a)$ ist, wobei

a

$a$ =

e

$e$ .

e^{u} \frac{d}{d x} [\sec (x) \ln (x)]

$e^{u} \frac{d}{d x} [\sec (x) \ln (x)]$

Schritt 2.3

Ersetze alle

u

$u$ durch

\sec (x) \ln (x)

$\sec (x) \ln (x)$ .

e^{\sec (x) \ln (x)} \frac{d}{d x} [\sec (x) \ln (x)]

$e^{\sec (x) \ln (x)} \frac{d}{d x} [\sec (x) \ln (x)]$

e^{\sec (x) \ln (x)} \frac{d}{d x} [\sec (x) \ln (x)]

$e^{\sec (x) \ln (x)} \frac{d}{d x} [\sec (x) \ln (x)]$

Schritt 3

Differenziere unter Anwendung der Produktregel, die besagt, dass

\frac{d}{d x} [f (x) g (x)]

$\frac{d}{d x} [f (x) g (x)]$ gleich

f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]

$f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ ist mit

f (x) = \sec (x)

$f (x) = \sec (x)$ und

g (x) = \ln (x)

$g (x) = \ln (x)$ .

e^{\sec (x) \ln (x)} (\sec (x) \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [\sec (x)])

$e^{\sec (x) \ln (x)} (\sec (x) \frac{d}{d x} [\ln (x)] + \ln (x) \frac{d}{d x} [\sec (x)])$

Schritt 4

Die Ableitung von

\ln (x)

$\ln (x)$ nach

x

$x$ ist

\frac{1}{x}

$\frac{1}{x}$ .

e^{\sec (x) \ln (x)} (\sec (x) \frac{1}{x} + \ln (x) \frac{d}{d x} [\sec (x)])

$e^{\sec (x) \ln (x)} (\sec (x) \frac{1}{x} + \ln (x) \frac{d}{d x} [\sec (x)])$

Schritt 5

Kombiniere

\sec (x)

$\sec (x)$ und

\frac{1}{x}

$\frac{1}{x}$ .

e^{\sec (x) \ln (x)} (\frac{\sec (x)}{x} + \ln (x) \frac{d}{d x} [\sec (x)])

$e^{\sec (x) \ln (x)} (\frac{\sec (x)}{x} + \ln (x) \frac{d}{d x} [\sec (x)])$

Schritt 6

Die Ableitung von

\sec (x)

$\sec (x)$ nach

x

$x$ ist

\sec (x) \tan (x)

$\sec (x) \tan (x)$ .

e^{\sec (x) \ln (x)} (\frac{\sec (x)}{x} + \ln (x) (\sec (x) \tan (x)))

$e^{\sec (x) \ln (x)} (\frac{\sec (x)}{x} + \ln (x) (\sec (x) \tan (x)))$

Schritt 7

Vereinfache.

Tippen, um mehr Schritte zu sehen ...

Schritt 7.1

Wende das Distributivgesetz an.

e^{\sec (x) \ln (x)} \frac{\sec (x)}{x} + e^{\sec (x) \ln (x)} (\ln (x) (\sec (x) \tan (x)))

$e^{\sec (x) \ln (x)} \frac{\sec (x)}{x} + e^{\sec (x) \ln (x)} (\ln (x) (\sec (x) \tan (x)))$

Schritt 7.2

Kombiniere

e^{\sec (x) \ln (x)}

$e^{\sec (x) \ln (x)}$ und

\frac{\sec (x)}{x}

$\frac{\sec (x)}{x}$ .

\frac{e^{\sec (x) \ln (x)} \sec (x)}{x} + e^{\sec (x) \ln (x)} \ln (x) \sec (x) \tan (x)

$\frac{e^{\sec (x) \ln (x)} \sec (x)}{x} + e^{\sec (x) \ln (x)} \ln (x) \sec (x) \tan (x)$

Schritt 7.3

Stelle die Terme um.

e^{\sec (x) \ln (x)} \sec (x) \tan (x) \ln (x) + \frac{e^{\sec (x) \ln (x)} \sec (x)}{x}

$e^{\sec (x) \ln (x)} \sec (x) \tan (x) \ln (x) + \frac{e^{\sec (x) \ln (x)} \sec (x)}{x}$

e^{\sec (x) \ln (x)} \sec (x) \tan (x) \ln (x) + \frac{e^{\sec (x) \ln (x)} \sec (x)}{x}

$e^{\sec (x) \ln (x)} \sec (x) \tan (x) \ln (x) + \frac{e^{\sec (x) \ln (x)} \sec (x)}{x}$