derivative f(x)=sqrt((x^2-3x)/(2x+1))

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

\int \frac{( 7 x ^{2} + x + 7 )}{( x ^{2} + 1 ) ^{2}} d x

\int \frac{x ^{3} + 1}{x ^{2} - 4} d x

\int_{0}^{4} e^{- x} d x

n = 1 \sum \infty (\frac{1}{4})^{n}

d er i v a t i v e f (x) = x^{4} + 5 x