derivative of (x^2+1\sqrt[3]{x^2+2})

Lösung

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

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

Schritte zur Lösung

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

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2} + 1, g = 3 x^{2} + 2

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

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

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

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

Multipliziere Brüche:

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

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

d er i v a t i v e - \frac{3}{x}

\int \frac{x + 4}{( x ^{2} + 8 x - 7 ) ^{2}} d x

\frac{d}{d t} (t + 2)

\int (3 x - 2)^{5} d x

d er i v a t i v e f (x) = \frac{3}{4 x}