derivative of sqrt(2x^2+4x+5)

Lösung

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

e^{x + 3 y} d y = x d x

\int \frac{8}{x ^{2} + 1} d x

\int \frac{x ^{2}}{x ^{2} + 2 x + 2} d x

\frac{\partial}{\partial x} (x + 2 y - 1)

a re a - x^{2} - 3 x, [- 6, - 1]