derivative of ta{g}(t,xx)

解

\frac{d}{d x} (t a g (t, x) x)

t a (\frac{d}{d x} (g (t, x)) x + 1 \cdot g (t, x))

解答ステップ

\frac{d}{d x} (t a g (t, x) x)

a, t

を定数として扱う

定数を除去:

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

= t a \frac{d}{d x} (g (t, x) x)

積の法則を適用:

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

f = g (t, x), g = x

= t a (\frac{d}{d x} (g (t, x)) x + \frac{d x}{d x} g (t, x))

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

= t a (\frac{d}{d x} (g (t, x)) x + 1 \cdot g (t, x))