derivative of ((x+1^3)/3+1/(4x+4))

解

\frac{d}{d x} (\frac{( x + 1 ) ^{3}}{3} + \frac{1}{4 x + 4})

\frac{4 x ^{4} + 16 x ^{3} + 20 x ^{2} + 8 x - 1}{4 ( x + 1 ) ^{2}} + 1

解答ステップ

\frac{d}{d x} (\frac{( x + 1 ) ^{3}}{3} + \frac{1}{4 x + 4})

和/差の法則を適用:

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

= \frac{d}{d x} (\frac{( x + 1 ) ^{3}}{3}) + \frac{d}{d x} (\frac{1}{4 x + 4})

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

\frac{d}{d x} (\frac{1}{4 x + 4}) = - \frac{1}{4 ( x + 1 ) ^{2}}

= (x + 1)^{2} - \frac{1}{4 ( x + 1 ) ^{2}}

簡素化

(x + 1)^{2} - \frac{1}{4 ( x + 1 ) ^{2}} : \frac{4 x ^{4} + 16 x ^{3} + 20 x ^{2} + 8 x - 1}{4 ( x + 1 ) ^{2}} + 1

= \frac{4 x ^{4} + 16 x ^{3} + 20 x ^{2} + 8 x - 1}{4 ( x + 1 ) ^{2}} + 1