derivative f(x)=(x+1)(x+1-1/(x+2))

解

導関数

f (x) = (x + 1) (x + 1 - \frac{1}{x + 2})

\frac{2 x ^{3} + 8 x ^{2} + 8 x - 1}{( x + 2 ) ^{2}} + 2

解答ステップ

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

積の法則を適用:

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

f = x + 1, g = x + 1 - \frac{1}{x + 2}

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

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

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

= 1 \cdot (x + 1 - \frac{1}{x + 2}) + (\frac{1}{( x + 2 ) ^{2}} + 1) (x + 1)

簡素化

1 \cdot (x + 1 - \frac{1}{x + 2}) + (\frac{1}{( x + 2 ) ^{2}} + 1) (x + 1) : \frac{2 x ^{3} + 8 x ^{2} + 8 x - 1}{( x + 2 ) ^{2}} + 2

= \frac{2 x ^{3} + 8 x ^{2} + 8 x - 1}{( x + 2 ) ^{2}} + 2