derivative of (2x^2-2xle{f}(x,t)t(x^2-3))

解

\frac{d}{d x} ((2 x^{2} - 2 x) l e f (x, t) t (x^{2} - 3))

l e t ((4 x - 2) f (x, t) (x^{2} - 3) + (\frac{d}{d x} (f (x, t)) (x^{2} - 3) + 2 x f (x, t)) (2 x^{2} - 2 x))

解答ステップ

\frac{d}{d x} ((2 x^{2} - 2 x) l e f (x, t) t (x^{2} - 3))

l, t

を定数として扱う

定数を除去:

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

= l e t \frac{d}{d x} ((2 x^{2} - 2 x) f (x, t) (x^{2} - 3))

積の法則を適用:

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

f = 2 x^{2} - 2 x, g = f (x, t) (x^{2} - 3)

= l e t (\frac{d}{d x} (2 x^{2} - 2 x) f (x, t) (x^{2} - 3) + \frac{d}{d x} (f (x, t) (x^{2} - 3)) (2 x^{2} - 2 x))

\frac{d}{d x} (2 x^{2} - 2 x) = 4 x - 2

\frac{d}{d x} (f (x, t) (x^{2} - 3)) = \frac{d}{d x} (f (x, t)) (x^{2} - 3) + 2 x f (x, t)

= l e t ((4 x - 2) f (x, t) (x^{2} - 3) + (\frac{d}{d x} (f (x, t)) (x^{2} - 3) + 2 x f (x, t)) (2 x^{2} - 2 x))