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

Lösung

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

- \frac{8 ( 3 x ^{2} - 1 )}{( x ^{3} - x + 1 ) ^{9}}

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

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

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

Vereinfache

4 (\frac{1}{( x ^{3} - x + 1 ) ^{2}})^{3} (- \frac{2 ( 3 x ^{2} - 1 )}{( x ^{3} - x + 1 ) ^{3}}) : - \frac{8 ( 3 x ^{2} - 1 )}{( x ^{3} - x + 1 ) ^{9}}

= - \frac{8 ( 3 x ^{2} - 1 )}{( x ^{3} - x + 1 ) ^{9}}

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

\frac{d}{d x} (16 - x^{\frac{2}{3}})

d er i v a t i v e cos ((π x)^{6})

\frac{d}{d x} (x x^{2} - 16)

in v erse l a pl a ce \frac{10}{s ^{2} + s + 10}