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

Lösung

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

\frac{3 x + 4}{4 ( x + 1 ) x + 1}

Schritte zur Lösung

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Vereinfache

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

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

(x + 1) y^{^{'}} = x + 6

\frac{\partial}{\partial x} (3 x^{4} - sin (5 x) + 4 x)

x \to - 2 lim (x + 2)

(y - x) y^{^{'}} = y - x + 8

x \to 0 lim (2 x + 1 - 1)