derivative of (x^3-3x^2+3x+1^'-(x-1)^')

Lösung

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

' (3 x^{2} - 6 x + 3) (x^{3} - 3 x^{2} + 3 x + 1)^{' - 1} -' (x - 1)^{' - 1}

Schritte zur Lösung

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

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d x} ((x - 1)^{^{'}}) =^{'} (x - 1)^{^{'} - 1}

=^{^{'}} (3 x^{2} - 6 x + 3) (x^{3} - 3 x^{2} + 3 x + 1)^{^{^{'}} - 1} -^{^{'}} (x - 1)^{^{^{'}} - 1}

d er i v a t i v e (x - 1)^{3} (x + 1)^{2}

d er i v a t i v e 6 (x^{2} + 3)^{- 1}

d er i v a t i v e - 9 sin^{3} (x)

d er i v a t i v e 12 x - 4

d er i v a t i v e 3 tan (x) - 4 cot (x_{7}) x^{2}