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

Lösung

ableitung von

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

\frac{- 18 x ^{2} - 8 x - 15}{( 2 x ^{2} - 5 ) ^{3}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (3 x + 1) = 3

\frac{d}{d x} ((2 x^{2} - 5)^{2}) = 8 x (2 x^{2} - 5)

= \frac{3 ( 2 x ^{2} - 5 ) ^{2} - 8 x ( 2 x ^{2} - 5 ) ( 3 x + 1 )}{( ( 2 x ^{2} - 5 ) ^{2} ) ^{2}}

Vereinfache

\frac{3 ( 2 x ^{2} - 5 ) ^{2} - 8 x ( 2 x ^{2} - 5 ) ( 3 x + 1 )}{( ( 2 x ^{2} - 5 ) ^{2} ) ^{2}} : \frac{- 18 x ^{2} - 8 x - 15}{( 2 x ^{2} - 5 ) ^{3}}

= \frac{- 18 x ^{2} - 8 x - 15}{( 2 x ^{2} - 5 ) ^{3}}

l a pl a ce t r an s f or m (1 + e^{t})^{2}

\frac{\partial}{\partial x} (8 + x ln (x y - 5))

\int \frac{x ^{2}}{u} d u

d er i v a t i v e f (x) = x^{2} arctan (4 x)

\frac{\partial}{\partial y} (e^{x} + y^{2})