derivative of (2x^3+5/(4x^2+7))

Lösung

\frac{d}{d x} (\frac{2 x ^{3} + 5}{4 x ^{2} + 7})

\frac{8 x ^{4} + 42 x ^{2} - 40 x}{( 4 x ^{2} + 7 ) ^{2}}

Schritte zur Lösung

\frac{d}{d x} (\frac{2 x ^{3} + 5}{4 x ^{2} + 7})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 2 x ^{3} + 5 ) ( 4 x ^{2} + 7 ) - \frac{d}{d x} ( 4 x ^{2} + 7 ) ( 2 x ^{3} + 5 )}{( 4 x ^{2} + 7 ) ^{2}}

\frac{d}{d x} (2 x^{3} + 5) = 6 x^{2}

\frac{d}{d x} (4 x^{2} + 7) = 8 x

= \frac{6 x ^{2} ( 4 x ^{2} + 7 ) - 8 x ( 2 x ^{3} + 5 )}{( 4 x ^{2} + 7 ) ^{2}}

Multipliziere aus

6 x^{2} (4 x^{2} + 7) - 8 x (2 x^{3} + 5) : 8 x^{4} + 42 x^{2} - 40 x

= \frac{8 x ^{4} + 42 x ^{2} - 40 x}{( 4 x ^{2} + 7 ) ^{2}}

\frac{d}{d x} ((f (x) (x^{3}))^{2})

\frac{d}{d x} (3 cot (2 x) sec (2 x))

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

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

e x p an d \frac{6}{( 1 - x ) ^{3}}