derivative of (4x^3+1/(x^2-x+3))

Lösung

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

\frac{4 x ^{4} - 8 x ^{3} + 36 x ^{2} - 2 x + 1}{( x ^{2} - x + 3 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (4 x^{3} + 1) = 12 x^{2}

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

= \frac{12 x ^{2} ( x ^{2} - x + 3 ) - ( 2 x - 1 ) ( 4 x ^{3} + 1 )}{( x ^{2} - x + 3 ) ^{2}}

Multipliziere aus

12 x^{2} (x^{2} - x + 3) - (2 x - 1) (4 x^{3} + 1) : 4 x^{4} - 8 x^{3} + 36 x^{2} - 2 x + 1

= \frac{4 x ^{4} - 8 x ^{3} + 36 x ^{2} - 2 x + 1}{( x ^{2} - x + 3 ) ^{2}}

\frac{d y}{d x} - \frac{y}{x} = \frac{x ^{2} + y ^{2}}{x}

d er i v a t i v e f (x) = e^{\frac{1}{2} x}

\frac{d}{d x} (x + 7)

im pl i c i t \frac{d y}{d x}, x^{2} + 5 x + 9 x y - y^{2} = 4

1 + 6 x^{\frac{3}{2}}