derivative of xlog_{e}(2x)

解

\frac{d}{d x} (x lo g_{e} (2 x))

ln (2 x) + 1

解答ステップ

\frac{d}{d x} (x lo g_{e} (2 x))

簡素化

x lo g_{e} (2 x) : x ln (2 x)

= \frac{d}{d x} (x ln (2 x))

積の法則を適用:

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

f = x, g = ln (2 x)

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

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

\frac{d}{d x} (ln (2 x)) = \frac{1}{x}

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

簡素化

1 \cdot ln (2 x) + \frac{1}{x} x : ln (2 x) + 1

= ln (2 x) + 1