derivative of (4x^3+9/x)

Lösung

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

\frac{8 x ^{3} - 9}{x ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (4 x^{3} + 9) = 12 x^{2}

\frac{d x}{d x} = 1

= \frac{12 x ^{2} x - 1 \cdot ( 4 x ^{3} + 9 )}{x ^{2}}

Vereinfache

\frac{12 x ^{2} x - 1 \cdot ( 4 x ^{3} + 9 )}{x ^{2}} : \frac{8 x ^{3} - 9}{x ^{2}}

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

\frac{d}{d t} (sin (ω t + θ))

x e^{x} y^{^{'}} + 2 e^{x} y = 0

\int 9 - 16 x^{2} d x

\int (\frac{1}{x ^{2} - 9}) d x

\frac{\partial}{\partial x} ((x^{2} + 9 x - 4 y)^{6})