derivative of 1/((3x^3+5x+4^{3/4)})

Lösung

\frac{d}{d x} (\frac{1}{( 3 x ^{3} + 5 x + 4 ) ^{\frac{3}{4}}})

- \frac{3 ( 9 x ^{2} + 5 )}{4 ( 3 x ^{3} + 5 x + 4 ) ^{\frac{7}{4}}}

Schritte zur Lösung

\frac{d}{d x} (\frac{1}{( 3 x ^{3} + 5 x + 4 ) ^{\frac{3}{4}}})

Wende Exponentenregel an:

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

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

Wende die Kettenregel an:

- \frac{3}{4 ( 3 x ^{3} + 5 x + 4 ) ^{\frac{7}{4}}} \frac{d}{d x} (3 x^{3} + 5 x + 4)

= - \frac{3}{4 ( 3 x ^{3} + 5 x + 4 ) ^{\frac{7}{4}}} \frac{d}{d x} (3 x^{3} + 5 x + 4)

\frac{d}{d x} (3 x^{3} + 5 x + 4) = 9 x^{2} + 5

= - \frac{3}{4 ( 3 x ^{3} + 5 x + 4 ) ^{\frac{7}{4}}} (9 x^{2} + 5)

Multipliziere Brüche:

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

= - \frac{3 ( 9 x ^{2} + 5 )}{4 ( 3 x ^{3} + 5 x + 4 ) ^{\frac{7}{4}}}

\frac{d}{d x} ((f (x) (6 x))^{2} + 3)

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

x \to 0 lim (\frac{x}{e ^{- 3 x} - 1})

3 x^{^{'}} + 6 x = 4, x (0) = 2

\frac{d}{d x} (1100 x^{2} - 0.2 x)