derivative f(x)=(2x+1)^5+10x(2x+1)^4

Lösung

ableitung von

f (x) = (2 x + 1)^{5} + 10 x (2 x + 1)^{4}

960 x^{4} + 1600 x^{3} + 960 x^{2} + 240 x + 20

Schritte zur Lösung

\frac{d}{d x} ((2 x + 1)^{5} + 10 x (2 x + 1)^{4})

Vereinfache

(2 x + 1)^{5} + 10 x (2 x + 1)^{4} : 192 x^{5} + 400 x^{4} + 320 x^{3} + 120 x^{2} + 20 x + 1

= \frac{d}{d x} (192 x^{5} + 400 x^{4} + 320 x^{3} + 120 x^{2} + 20 x + 1)

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (192 x^{5}) + \frac{d}{d x} (400 x^{4}) + \frac{d}{d x} (320 x^{3}) + \frac{d}{d x} (120 x^{2}) + \frac{d}{d x} (20 x) + \frac{d}{d x} (1)

\frac{d}{d x} (192 x^{5}) = 960 x^{4}

\frac{d}{d x} (400 x^{4}) = 1600 x^{3}

\frac{d}{d x} (320 x^{3}) = 960 x^{2}

\frac{d}{d x} (120 x^{2}) = 240 x

\frac{d}{d x} (20 x) = 20

\frac{d}{d x} (1) = 0

= 960 x^{4} + 1600 x^{3} + 960 x^{2} + 240 x + 20 + 0

Vereinfache

= 960 x^{4} + 1600 x^{3} + 960 x^{2} + 240 x + 20

y^{^{'}} + \frac{y ^{2}}{x ^{2}} = 0

\frac{d}{d x} (\frac{x ^{2} - x}{x})

\frac{d y}{d x} + 4 y = - x

\int \frac{4}{- 3 x} d x

\frac{d}{d x} (I n^{2})