derivative of (8x^3-9/(x^2))

Lösung

\frac{d}{d x} (\frac{8 x ^{3} - 9}{x ^{2}})

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

Schritte zur Lösung

\frac{d}{d x} (\frac{8 x ^{3} - 9}{x ^{2}})

Wende die Quotientenregel an:

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

= \frac{\frac{d}{d x} ( 8 x ^{3} - 9 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( 8 x ^{3} - 9 )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (8 x^{3} - 9) = 24 x^{2}

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

= \frac{24 x ^{2} x ^{2} - 2 x ( 8 x ^{3} - 9 )}{( x ^{2} ) ^{2}}

Vereinfache

\frac{24 x ^{2} x ^{2} - 2 x ( 8 x ^{3} - 9 )}{( x ^{2} ) ^{2}} : \frac{2 ( 4 x ^{3} + 9 )}{x ^{3}}

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

\int \frac{1}{u ^{5}} d u

x^{2} y^{^{''}} - 3 x y^{^{'}} + 3 y = 0, y (1) = 5, y^{^{'}} (1) = 3

x (e^{\frac{y}{x}} - 1) y^{^{'}} = e^{\frac{y}{x}} (y - x)

d er i v a t i v e y = ln (x^{5}) + e^{x} - ln (e^{2})

d er i v a t i v e y = (3 x + 1)^{2}