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

Lösung

ableitung von

f (x) = \frac{π ^{2}}{x ^{2} + 4 x}

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

Schritte zur Lösung

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

Entferne die Konstante:

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

= π^{2} \frac{d}{d x} (\frac{1}{x ^{2} + 4 x})

Wende Exponentenregel an:

\frac{1}{a} = a^{- 1}

= π^{2} \frac{d}{d x} ((x^{2} + 4 x)^{- 1})

Wende die Kettenregel an:

- \frac{1}{( x ^{2} + 4 x ) ^{2}} \frac{d}{d x} (x^{2} + 4 x)

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

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

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

Vereinfache

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

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

\int - \frac{1}{27} sin (3 x) d x

\int \frac{3 - 5 x}{9 - 5 x ^{2}} d x

\frac{d}{d x} (arccos (\frac{6}{x}))

\frac{\partial}{\partial v} (e^{u (v)} + sin (v))

\int_{1}^{2} (- y^{3} + 2 y^{2} + 5 y - 6) d y