derivative of (3x^2-1/(xsqrt(1-x^2)))

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (3 x^{2} - 1) = 6 x

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

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

Vereinfache

\frac{6 xx 1 - x ^{2} - \frac{- 2 x ^{2} + 1}{1 - x ^{2}} ( 3 x ^{2} - 1 )}{( x 1 - x ^{2} ) ^{2}} : \frac{x ^{2} + 1}{x ^{2} ( 1 - x ^{2} ) 1 - x ^{2}}

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

t an g e n t f (x) = 9 x - 9 x

t an g e n t f (x) = - 3 cos (x), at x = \frac{- π}{3}

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

s im pl i f y \frac{- 3 x ^{5} - x ^{4} - x ^{3} + 4 x ^{2}}{x ^{2}}

\int x (x^{3} - \frac{1}{x}) d x