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

Lösung

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

\frac{x ^{4} + 9 x ^{2} - 2 x}{( x ^{2} + 3 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Multipliziere aus

3 x^{2} (x^{2} + 3) - 2 x (x^{3} + 1) : x^{4} + 9 x^{2} - 2 x

= \frac{x ^{4} + 9 x ^{2} - 2 x}{( x ^{2} + 3 ) ^{2}}

d er i v a t i v e (sin (x) + 2 tan (x)) (sec (x) - 3 csc (x))

\frac{d}{d x} (\frac{1}{x ^{12}})

d er i v a t i v e tan (2 x^{2} - x + 3)

d er i v a t i v e y = 9 sec (\frac{1}{3} x)

f (x) = 2 x ln (x) + x