derivative of (x^2-x+1/(x^2+1))

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Multipliziere aus

(2 x - 1) (x^{2} + 1) - 2 x (x^{2} - x + 1) : x^{2} - 1

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

\frac{d y}{d x} + \frac{3}{3.5} y = 9.8

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

\frac{\partial}{\partial x} (sin (x) + x cos (x))

\frac{d y}{d t} = 4 - \frac{1.5 y}{10 + 2.5 t}

\int 6 tan^{4} (x) d x