derivative ln(x^2-4x+1)

Lösung

ableitung von

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

\frac{2 x - 4}{x ^{2} - 4 x + 1}

Schritte zur Lösung

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

Wende die Kettenregel an:

\frac{1}{x ^{2} - 4 x + 1} \frac{d}{d x} (x^{2} - 4 x + 1)

= \frac{1}{x ^{2} - 4 x + 1} \frac{d}{d x} (x^{2} - 4 x + 1)

\frac{d}{d x} (x^{2} - 4 x + 1) = 2 x - 4

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

Vereinfache

\frac{1}{x ^{2} - 4 x + 1} (2 x - 4) : \frac{2 x - 4}{x ^{2} - 4 x + 1}

= \frac{2 x - 4}{x ^{2} - 4 x + 1}

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

x \to 0 lim (\frac{( 3 x + 8 - 2 )}{x})

\int 5 x \cdot cos (8 x) d x

\frac{d}{d x} (- sin^{2} (x))

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