derivative of ln((3x/(x^2+4)))

Lösung

\frac{d}{d x} (ln (\frac{3 x}{x ^{2} + 4}))

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

Schritte zur Lösung

\frac{d}{d x} (ln (\frac{3 x}{x ^{2} + 4}))

Wende die Kettenregel an:

\frac{1}{\frac{3 x}{x ^{2} + 4}} \frac{d}{d x} (\frac{3 x}{x ^{2} + 4})

= \frac{1}{\frac{3 x}{x ^{2} + 4}} \frac{d}{d x} (\frac{3 x}{x ^{2} + 4})

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

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

Vereinfache

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

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

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

\frac{d y}{d x} = 8 x^{3} e^{- 2 y}, y (1) = 0

\frac{d}{d x} (\frac{x ^{2} + 14}{x ^{8}})

y^{^{''}} + 4 y^{^{'}} + 6.25 y = 3.125 (x + 1)^{2}

y^{^{'}} = 2 x y^{- 2}