derivative of (4x+3^4(x+1)^{-3})

解

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

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

解答ステップ

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

積の法則を適用:

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

f = (4 x + 3)^{4}, g = (x + 1)^{- 3}

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

\frac{d}{d x} ((4 x + 3)^{4}) = 16 (4 x + 3)^{3}

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

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

簡素化

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

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