derivative of (x^3/(sqrt(x)+x))

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

\frac{3 x ^{2} ( x + x ) - ( \frac{1}{2 x} + 1 ) x ^{3}}{( x + x ) ^{2}} : \frac{4 x ^{3} + 6 x ^{2} x - x ^{\frac{5}{2}}}{2 ( x + x ) ^{2}}

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

\frac{\partial}{\partial x} (1 + 2 x y)

in v erse l a pl a ce \frac{s}{( s + 2 ) ^{2} + 1}

y^{^{''}} - y^{^{'}} + 6 y = 0

\int_{2}^{\infty} \frac{1}{v ^{2} + 5 v - 6} d v

\frac{d}{d x} (ln (7 x^{2} + 9 y^{2}))