derivative of ln(((2x+1^3)/((3x-1)^4)))

Lösung

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

\frac{6 ( - x - 3 )}{( 2 x + 1 ) ( 3 x - 1 )}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

= \frac{6 ( - x - 3 )}{( 2 x + 1 ) ( 3 x - 1 )}

t a y l or z (z + 1) e^{z}

\frac{\partial}{\partial x} ((2 x y - 1) e^{- x y})

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

t an g e n t y = 8 x + 12 - 15 e^{x}, (0, - 3)

\int \frac{3 + x}{1 + x ^{2}} d x