derivative of xln(2x+1)

Lösung

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

ln (2 x + 1) + \frac{2 x}{2 x + 1}

Schritte zur Lösung

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

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = ln (2 x + 1)

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

\frac{d x}{d x} = 1

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

= 1 \cdot ln (2 x + 1) + \frac{2}{2 x + 1} x

Vereinfache

1 \cdot ln (2 x + 1) + \frac{2}{2 x + 1} x : ln (2 x + 1) + \frac{2 x}{2 x + 1}

= ln (2 x + 1) + \frac{2 x}{2 x + 1}

\frac{d}{d x} (\frac{1}{3} cos (3 x))

\frac{d}{d x} ((2 x - 4)^{4} (x^{2} + x + 1)^{5})

\int \frac{( 5 x + 3 )}{( x ^{2} + 4 x + 7 )} d x

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

k = 1 \sum \infty \frac{k}{k ^{2} + 1}