derivative of (3x^2-5/(2x^2+1))

Lösung

\frac{d}{d x} (\frac{3 x ^{2} - 5}{2 x ^{2} + 1})

\frac{26 x}{( 2 x ^{2} + 1 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{3 x ^{2} - 5}{2 x ^{2} + 1})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 3 x ^{2} - 5 ) ( 2 x ^{2} + 1 ) - \frac{d}{d x} ( 2 x ^{2} + 1 ) ( 3 x ^{2} - 5 )}{( 2 x ^{2} + 1 ) ^{2}}

\frac{d}{d x} (3 x^{2} - 5) = 6 x

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

= \frac{6 x ( 2 x ^{2} + 1 ) - 4 x ( 3 x ^{2} - 5 )}{( 2 x ^{2} + 1 ) ^{2}}

Multipliziere aus

6 x (2 x^{2} + 1) - 4 x (3 x^{2} - 5) : 26 x

= \frac{26 x}{( 2 x ^{2} + 1 ) ^{2}}

x \to - 8 + lim (\frac{( x + 7 )}{x + 8})

\frac{d}{d x} (e^{- a ∣ x ∣})

\int \frac{2 x}{( x ^{2} + 1 ) ^{90}} d x

d er i v a t i v e f (8) = 2 (x - 5)

h \to c lim (h^{3} + 4 h^{2} - 3)