derivative 1/((2x^3+3x+1)^{3/4)}

Lösung

ableitung von

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

- \frac{3 ( 6 x ^{2} + 3 )}{4 ( 2 x ^{3} + 3 x + 1 ) ^{\frac{7}{4}}}

Schritte zur Lösung

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

Wende Exponentenregel an:

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

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

Wende die Kettenregel an:

- \frac{3}{4 ( 2 x ^{3} + 3 x + 1 ) ^{\frac{7}{4}}} \frac{d}{d x} (2 x^{3} + 3 x + 1)

= - \frac{3}{4 ( 2 x ^{3} + 3 x + 1 ) ^{\frac{7}{4}}} \frac{d}{d x} (2 x^{3} + 3 x + 1)

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

= - \frac{3}{4 ( 2 x ^{3} + 3 x + 1 ) ^{\frac{7}{4}}} (6 x^{2} + 3)

Multipliziere Brüche:

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

= - \frac{3 ( 6 x ^{2} + 3 )}{4 ( 2 x ^{3} + 3 x + 1 ) ^{\frac{7}{4}}}

n or ma l y = e^{2 x}, at x = ln (2)

\frac{\partial}{\partial y} (y cos (x y) + e^{x})

d er i v a t i v e 3 3 x^{4} + 4 x^{3}

\int cos^{2} (t) - sin^{2} (t) d t

\frac{\partial}{\partial x} (sin (x^{1} y^{3}))