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

Lösung

\frac{d}{d 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}}}

3 y^{^{''}} + y = 0

\int_{0}^{1} 2 x \cdot 1 + 4 x^{2} d x

\int \frac{( 7 - x )}{( x ^{2} - 4 x + 4 )} d x

\int 5 x cos (4 x) d x

\frac{d}{d x} (\frac{cos ( x )}{1 - sin ( x )})