derivative (x^2)/(sqrt(x^2+5))

Lösung

ableitung von

\frac{x ^{2}}{x ^{2} + 5}

\frac{x ^{3} + 10 x}{( x ^{2} + 5 ) x ^{2} + 5}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

\frac{2 x x ^{2} + 5 - \frac{x}{x ^{2} + 5} x ^{2}}{( x ^{2} + 5 ) ^{2}} : \frac{x ^{3} + 10 x}{( x ^{2} + 5 ) x ^{2} + 5}

= \frac{x ^{3} + 10 x}{( x ^{2} + 5 ) x ^{2} + 5}

in v erse l a pl a ce \frac{1}{( s ^{4} - 1 )}

\int \frac{1}{7 + 5 x} d x

t an g e n t f (x) = \frac{x - 1}{x + 1}, at x = 0

t an g e n t x^{2} + x y - y^{2} = - 4, (2, 4)

y^{^{'}} = \frac{5 x ^{5} e ^{\frac{y}{x}} + x ^{3} y ^{2}}{x ^{4} y}