derivative of (|x-1|/(|x-3|))

Lösung

\frac{d}{d x} (\frac{∣ x - 1 ∣}{∣ x - 3 ∣})

- \frac{2 ( x - 1 )}{∣ x - 1 ∣ ∣ x - 3 ∣ ( x - 3 )}

Schritte zur Lösung

\frac{d}{d x} (\frac{∣ x - 1 ∣}{∣ x - 3 ∣})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( ∣ x - 1 ∣ ) ∣ x - 3 ∣ - \frac{d}{d x} ( ∣ x - 3 ∣ ) ∣ x - 1 ∣}{∣ x - 3 ∣ ^{2}}

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

\frac{d}{d x} (∣ x - 3 ∣) = \frac{x - 3}{∣ x - 3 ∣}

= \frac{\frac{x - 1}{∣ x - 1 ∣} ∣ x - 3 ∣ - \frac{x - 3}{∣ x - 3 ∣} ∣ x - 1 ∣}{∣ x - 3 ∣ ^{2}}

Vereinfache

\frac{\frac{x - 1}{∣ x - 1 ∣} ∣ x - 3 ∣ - \frac{x - 3}{∣ x - 3 ∣} ∣ x - 1 ∣}{∣ x - 3 ∣ ^{2}} : - \frac{2 ( x - 1 )}{∣ x - 1 ∣ ∣ x - 3 ∣ ( x - 3 )}

= - \frac{2 ( x - 1 )}{∣ x - 1 ∣ ∣ x - 3 ∣ ( x - 3 )}

t a y l or x ln (x)

d er i v a t i v e f (x) = \frac{x - 3}{5 x + 2}

t an g e n t - 4 cos (x), at x = \frac{π}{2}

\int \frac{4 x ^{2}}{x ^{2} + 8} d x

t an g e n t y = (\frac{( x - 1 )}{( x - 2 )})^{2}