derivative of (x^2-5x+4/(|x-1|))

Lösung

\frac{d}{d x} (\frac{x ^{2} - 5 x + 4}{∣ x - 1 ∣})

\frac{x - 1}{∣ x - 1 ∣}

Schritte zur Lösung

\frac{d}{d x} (\frac{x ^{2} - 5 x + 4}{∣ x - 1 ∣})

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d x} ( x ^{2} - 5 x + 4 ) ∣ x - 1 ∣ - \frac{d}{d x} ( ∣ x - 1 ∣ ) ( x ^{2} - 5 x + 4 )}{∣ x - 1 ∣ ^{2}}

\frac{d}{d x} (x^{2} - 5 x + 4) = 2 x - 5

\frac{d}{d x} (∣ x - 1 ∣) = \frac{x - 1}{∣ x - 1 ∣}

= \frac{( 2 x - 5 ) ∣ x - 1 ∣ - \frac{x - 1}{∣ x - 1 ∣} ( x ^{2} - 5 x + 4 )}{∣ x - 1 ∣ ^{2}}

Vereinfache

\frac{( 2 x - 5 ) ∣ x - 1 ∣ - \frac{x - 1}{∣ x - 1 ∣} ( x ^{2} - 5 x + 4 )}{∣ x - 1 ∣ ^{2}} : \frac{x - 1}{∣ x - 1 ∣}

= \frac{x - 1}{∣ x - 1 ∣}

\frac{d}{d x} (x^{1 - a})

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

x \to 1 lim (\frac{e ^{x - 1} - 1}{ln ^{2} ( x )})

a re a x^{2} - 4 x + 3, 3 + 4 x - x^{2}

\int_{- 1}^{1} (x^{2} - 1) d x