erweitern f''(x)(2x+1)^5

Lösung

erweitern

f^{''} (x) (2 x + 1)^{5}

32 x^{5} f'' (x) + 80 x^{4} f'' (x) + 80 x^{3} f'' (x) + 40 x^{2} f'' (x) + 10 x f'' (x) + f'' (x)

Schritte zur Lösung

f^{''} (x) (2 x + 1)^{5}

(2 x + 1)^{5} = 32 x^{5} + 80 x^{4} + 80 x^{3} + 40 x^{2} + 10 x + 1

= f^{^{''}} (x) (32 x^{5} + 80 x^{4} + 80 x^{3} + 40 x^{2} + 10 x + 1)

Setze Klammern

= f^{^{''}} (x) 32 x^{5} + f^{^{''}} (x) 80 x^{4} + f^{^{''}} (x) 80 x^{3} + f^{^{''}} (x) 40 x^{2} + f^{^{''}} (x) 10 x + f^{^{''}} (x) 1

= 32 x^{5} f^{^{''}} (x) + 80 x^{4} f^{^{''}} (x) + 80 x^{3} f^{^{''}} (x) + 40 x^{2} f^{^{''}} (x) + 10 x f^{^{''}} (x) + 1 f^{^{''}} (x)

Multipliziere:

1 f^{''} (x) = f^{''} (x)

= 32 x^{5} f^{^{''}} (x) + 80 x^{4} f^{^{''}} (x) + 80 x^{3} f^{^{''}} (x) + 40 x^{2} f^{^{''}} (x) + 10 x f^{^{''}} (x) + f^{^{''}} (x)

e x p an d 3 \frac{x + 2}{x ^{4} ( x ^{2} - 4 )}

e x p an d \frac{1}{4} x^{2} + \frac{3}{5} x + \frac{9}{5}

e x p an d 7 m^{2} n^{6} \cdot n^{4}

e x p an d - \frac{3}{( x + 1 )} + \frac{2}{x + 4}

e x p an d \frac{( 3 x + 2 ) ( 2 x - 1 )}{( 2 x - 2 ) ( 3 x + 1 )}