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

Lösung

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

40 x^{4} + 80 x^{3} + 24 x^{2}

Schritte zur Lösung

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

Wende die Produktregel an:

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

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

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

\frac{d}{d x} (2 x^{3} + 4 x^{2}) = 6 x^{2} + 8 x

\frac{d}{d x} (4 x^{2} + 2 x) = 8 x + 2

= (6 x^{2} + 8 x) (4 x^{2} + 2 x) + (8 x + 2) (2 x^{3} + 4 x^{2})

Vereinfache

(6 x^{2} + 8 x) (4 x^{2} + 2 x) + (8 x + 2) (2 x^{3} + 4 x^{2}) : 40 x^{4} + 80 x^{3} + 24 x^{2}

= 40 x^{4} + 80 x^{3} + 24 x^{2}

\int 225 - t^{2} d t

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

\frac{\partial}{\partial x} (\frac{x}{x + y ^{2}})

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = 0

3 r = \frac{d r}{d θ} - θ^{3}