derivative of (8x^2+2x+4/(sqrt(x)))

Lösung

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

\frac{12 x ^{2} + x - 2}{x x}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (8 x^{2} + 2 x + 4) = 16 x + 2

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

= \frac{( 16 x + 2 ) x - \frac{1}{2 x} ( 8 x ^{2} + 2 x + 4 )}{( x ) ^{2}}

Vereinfache

\frac{( 16 x + 2 ) x - \frac{1}{2 x} ( 8 x ^{2} + 2 x + 4 )}{( x ) ^{2}} : \frac{12 x ^{2} + x - 2}{x x}

= \frac{12 x ^{2} + x - 2}{x x}

\int 9 x^{2} ln (x) d x

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

x \frac{d y}{d x} + (3 x + 1) y = e^{3 x}

\int \frac{x}{3 - 2 x - x ^{2}} d x

\int_{2}^{\frac{1}{x}} arctan (t) d t