derivative f(x)=2xe^xcsc(x)

解

導関数

f (x) = 2 x e^{x} csc (x)

2 (e^{x} csc (x) + x (e^{x} csc (x) - e^{x} cot (x) csc (x)))

解答ステップ

\frac{d}{d x} (2 x e^{x} csc (x))

定数を除去:

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

= 2 \frac{d}{d x} (x e^{x} csc (x))

積の法則を適用:

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

f = x, g = e^{x} csc (x)

= 2 (\frac{d x}{d x} e^{x} csc (x) + \frac{d}{d x} (e^{x} csc (x)) x)

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

\frac{d}{d x} (e^{x} csc (x)) = e^{x} csc (x) - e^{x} cot (x) csc (x)

= 2 (1 \cdot e^{x} csc (x) + (e^{x} csc (x) - e^{x} cot (x) csc (x)) x)

簡素化

= 2 (e^{x} csc (x) + x (e^{x} csc (x) - e^{x} cot (x) csc (x)))