derivative of 0.6x^34^x

解

\frac{d}{d x} (0.6 x^{3} 4^{x})

0.6 (3 \cdot 4^{x} x^{2} + 0.69314 \dots \cdot 2^{2 x + 1} x^{3})

解答ステップ

\frac{d}{d x} (0.6 x^{3} \cdot 4^{x})

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.6 \frac{d}{d x} (x^{3} \cdot 4^{x})

積の法則を適用:

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

f = x^{3}, g = 4^{x}

= 0.6 (\frac{d}{d x} (x^{3}) 4^{x} + \frac{d}{d x} (4^{x}) x^{3})

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

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

= 0.6 (3 x^{2} \cdot 4^{x} + ln (2) \cdot 2^{2 x + 1} x^{3})

10進法形式に改善する

= 0.6 (3 \cdot 4^{x} x^{2} + 0.69314 \dots \cdot 2^{2 x + 1} x^{3})