derivative of ln^4(x^3+2x-e^{-3x})

Lösung

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

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Multipliziere Brüche:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

t an g e n t x^{3} - 25 x + 5, at 5

\frac{d}{d x} (3 tan (4 x))

\int \frac{x - 39}{x ^{2} - x - 12} d x

\int_{0}^{1} \frac{4 t ^{3}}{1 + t ^{4}} d t

y^{^{''}} + 9 y = 0, y (0) = - 6, y^{^{'}} (0) = - 9