erweitern 3(x^2+2y)^2(2x+2y'(x))

Lösung

erweitern

3 (x^{2} + 2 y)^{2} (2 x + 2 y^{'} (x))

6 x^{5} + 24 x^{3} y + 24 x y^{2} + 6 x^{4} y' (x) + 24 x^{2} y' (x) y + 24 y' (x) y^{2}

Schritte zur Lösung

3 (x^{2} + 2 y)^{2} (2 x + 2 y^{'} (x))

Wende das Distributivgesetz an:

a (b + c) = ab + a c

a = 3 (x^{2} + 2 y)^{2}, b = 2 x, c = 2 y^{'} (x)

= 3 (x^{2} + 2 y)^{2} \cdot 2 x + 3 (x^{2} + 2 y)^{2} \cdot 2 y^{^{'}} (x)

= 3 \cdot 2 x (x^{2} + 2 y)^{2} + 3 \cdot 2 (x^{2} + 2 y)^{2} y^{^{'}} (x)

Vereinfache

3 \cdot 2 x (x^{2} + 2 y)^{2} + 3 \cdot 2 (x^{2} + 2 y)^{2} y^{'} (x) : 6 x^{5} + 24 x^{3} y + 24 x y^{2} + 6 x^{4} y^{'} (x) + 24 x^{2} y^{'} (x) y + 24 y^{'} (x) y^{2}

= 6 x^{5} + 24 x^{3} y + 24 x y^{2} + 6 x^{4} y^{'} (x) + 24 x^{2} y^{'} (x) y + 24 y^{'} (x) y^{2}

e x p an d x (v^{2} - 1) d v + v^{3} - 3 v

e x p an d \frac{x ^{2} - 1}{( x - 2 ) ^{2}}

e x p an d (a + b x + c x^{2}) \cdot (d + e x + f x^{2})

e x p an d 5 (- 2 x - y = - 6)

\frac{d}{d y} ((q) (q^{0.3} + y^{0.3}))