derivative f(x)=(2x)/(x^2+4)

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Quotientenregel an:

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

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

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

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

= 2 \cdot \frac{1 \cdot ( x ^{2} + 4 ) - 2 xx}{( x ^{2} + 4 ) ^{2}}

Vereinfache

2 \cdot \frac{1 \cdot ( x ^{2} + 4 ) - 2 xx}{( x ^{2} + 4 ) ^{2}} : \frac{2 ( - x ^{2} + 4 )}{( x ^{2} + 4 ) ^{2}}

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

\int - 9 x^{2} sec (3 x^{3} + 1) tan (3 x^{3} + 1) d x

\int in x d x

\frac{\partial}{\partial x} (y f (x) + 2 x^{3})

d er i v a t i v e e^{1 - x} \cdot (5 - x^{2})

d er i v a t i v e f (x) = (7 x^{4} - 6 x^{3} - 5 x)^{8}