derivative of 2^{sin(x}*ta{g}(x,t)(x-x^3))

解

\frac{d}{d x} (2^{s i n (x)} \cdot t a g (x, t) (x - x^{3}))

t a (ln (2) \cdot 2^{s i n (x)} cos (x) g (x, t) (x - x^{3}) + (\frac{d}{d x} (g (x, t)) (x - x^{3}) + (1 - 3 x^{2}) g (x, t)) \cdot 2^{s i n (x)})

解答ステップ

\frac{d}{d x} (2^{s i n (x)} t a g (x, t) (x - x^{3}))

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} (2^{s i n (x)} g (x, t) (x - x^{3}))

積の法則を適用:

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

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

= t a (\frac{d}{d x} (2^{s i n (x)}) g (x, t) (x - x^{3}) + \frac{d}{d x} (g (x, t) (x - x^{3})) 2^{s i n (x)})

\frac{d}{d x} (2^{s i n (x)}) = ln (2) \cdot 2^{s i n (x)} cos (x)

\frac{d}{d x} (g (x, t) (x - x^{3})) = \frac{d}{d x} (g (x, t)) (x - x^{3}) + (1 - 3 x^{2}) g (x, t)

= t a (ln (2) \cdot 2^{s i n (x)} cos (x) g (x, t) (x - x^{3}) + (\frac{d}{d x} (g (x, t)) (x - x^{3}) + (1 - 3 x^{2}) g (x, t)) \cdot 2^{s i n (x)})