derivative of sqrt(3x^2+1)(3x^4+1^3)

Lösung

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

\frac{3 x ( 3 x ^{4} + 1 ) ^{3}}{3 x ^{2} + 1} + 36 x^{3} (3 x^{4} + 1)^{2} 3 x^{2} + 1

Schritte zur Lösung

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

Wende die Produktregel an:

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

f = 3 x^{2} + 1, g = (3 x^{4} + 1)^{3}

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

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

\frac{d}{d x} ((3 x^{4} + 1)^{3}) = 36 x^{3} (3 x^{4} + 1)^{2}

= \frac{3 x}{3 x ^{2} + 1} (3 x^{4} + 1)^{3} + 36 x^{3} (3 x^{4} + 1)^{2} 3 x^{2} + 1

Multipliziere Brüche:

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

= \frac{3 x ( 3 x ^{4} + 1 ) ^{3}}{3 x ^{2} + 1} + 36 x^{3} (3 x^{4} + 1)^{2} 3 x^{2} + 1

\int x^{3} cos (x^{4} - 5) d x

t \to \infty lim (e^{- t})

d er i v a t i v e 2 x^{\frac{1}{3}}

\int (2 x + 1) sin (x) d x

\frac{d}{d x} (a^{2} x + a^{2} y + b^{2} x + b^{2} y)