derivative f(x)=sqrt((4x+1)/(2x^2+1))

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

\frac{d}{d x} (\frac{4 x + 1}{2 x ^{2} + 1}) = \frac{- 8 x ^{2} - 4 x + 4}{( 2 x ^{2} + 1 ) ^{2}}

= \frac{1}{2 \frac{4 x + 1}{2 x ^{2} + 1}} \cdot \frac{- 8 x ^{2} - 4 x + 4}{( 2 x ^{2} + 1 ) ^{2}}

Vereinfache

\frac{1}{2 \frac{4 x + 1}{2 x ^{2} + 1}} \cdot \frac{- 8 x ^{2} - 4 x + 4}{( 2 x ^{2} + 1 ) ^{2}} : \frac{2 ( - 2 x ^{2} - x + 1 )}{4 x + 1 ( 2 x ^{2} + 1 ) ^{\frac{3}{2}}}

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

\int \frac{x - y}{x + y} d y

\int 9 x^{\frac{1}{2}} d x

x \to \infty lim (2 x - 3)

t a y l or \frac{5}{x}, 4

n = 0 \sum \infty \frac{1}{e ^{5 n}}