derivative of (xcos(x+pi)/(x-pi))

Lösung

\frac{d}{d x} (\frac{x cos ( x ) + π}{x - π})

\frac{- x ^{2} sin ( x ) + π x sin ( x ) - π cos ( x ) - π}{( x - π ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{x cos ( x ) + π}{x - π})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( x cos ( x ) + π ) ( x - π ) - \frac{d}{d x} ( x - π ) ( x cos ( x ) + π )}{( x - π ) ^{2}}

\frac{d}{d x} (x cos (x) + π) = - x sin (x) + cos (x)

\frac{d}{d x} (x - π) = 1

= \frac{( - x sin ( x ) + cos ( x ) ) ( x - π ) - 1 \cdot ( x cos ( x ) + π )}{( x - π ) ^{2}}

Vereinfache

\frac{( - x sin ( x ) + cos ( x ) ) ( x - π ) - 1 \cdot ( x cos ( x ) + π )}{( x - π ) ^{2}} : \frac{- x ^{2} sin ( x ) + π x sin ( x ) - π cos ( x ) - π}{( x - π ) ^{2}}

= \frac{- x ^{2} sin ( x ) + π x sin ( x ) - π cos ( x ) - π}{( x - π ) ^{2}}

x \to 0 lim ((1 - e^{4 x})^{x})

\frac{d}{d x} (sin^{3} (cos (2 x)))

\int \frac{1}{25 + x ^{2}} d x

(1 + y) d y = 2 t d t

\int \frac{x ^{2}}{16 - x ^{2}} d x