derivative of 2xtan(2x+2)

解

\frac{d}{d x} (2 x tan (2 x + 2))

2 (tan (2 x + 2) + 2 x sec^{2} (2 x + 2))

解答ステップ

\frac{d}{d x} (2 x tan (2 x + 2))

定数を除去:

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

= 2 \frac{d}{d x} (x tan (2 x + 2))

積の法則を適用:

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

f = x, g = tan (2 x + 2)

= 2 (\frac{d x}{d x} tan (2 x + 2) + \frac{d}{d x} (tan (2 x + 2)) x)

\frac{d x}{d x} = 1

\frac{d}{d x} (tan (2 x + 2)) = sec^{2} (2 x + 2) \cdot 2

= 2 (1 \cdot tan (2 x + 2) + sec^{2} (2 x + 2) \cdot 2 x)

簡素化

= 2 (tan (2 x + 2) + 2 x sec^{2} (2 x + 2))