derivative f(x)=(x+2)^{x^2-2x+1}

Lösung

ableitung von

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

(x^{2} - 2 x + 1 + 2 x^{2} ln (x + 2) + 2 x ln (x + 2) - 4 ln (x + 2)) (x + 2)^{x^{2} - 2 x}

Schritte zur Lösung

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

Wende Exponentenregel an:

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

(x + 2)^{x^{2} - 2 x + 1} = e^{(x^{2} - 2 x + 1) l n (x + 2)}

= \frac{d}{d x} (e^{(x^{2} - 2 x + 1) l n (x + 2)})

Wende die Kettenregel an:

e^{(x^{2} - 2 x + 1) l n (x + 2)} \frac{d}{d x} ((x^{2} - 2 x + 1) ln (x + 2))

= e^{(x^{2} - 2 x + 1) l n (x + 2)} \frac{d}{d x} ((x^{2} - 2 x + 1) ln (x + 2))

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

= e^{(x^{2} - 2 x + 1) l n (x + 2)} ((2 x - 2) ln (x + 2) + \frac{1}{x + 2} (x^{2} - 2 x + 1))

Vereinfache

e^{(x^{2} - 2 x + 1) l n (x + 2)} ((2 x - 2) ln (x + 2) + \frac{1}{x + 2} (x^{2} - 2 x + 1)) : (x^{2} - 2 x + 1 + 2 x^{2} ln (x + 2) + 2 x ln (x + 2) - 4 ln (x + 2)) (x + 2)^{x^{2} - 2 x}

= (x^{2} - 2 x + 1 + 2 x^{2} ln (x + 2) + 2 x ln (x + 2) - 4 ln (x + 2)) (x + 2)^{x^{2} - 2 x}

\frac{d}{d x} (2 3 x - \frac{7}{x})

\int_{3}^{5} \frac{x}{x ^{2} - 2} d x

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

\int 4 x^{2} sin (π x) d x

\frac{\partial}{\partial x} (ln (2 x^{2} y))