derivative f(x)=4^{sin(2)(x)+cos(3x)}

解

導関数

f (x) = 4^{s i n (2) (x) + c o s (3 x)}

ln (2) \cdot 2^{2 (s i n (2) x + c o s (3 x)) + 1} (sin (2) - 3 sin (3 x))

解答ステップ

\frac{d}{d x} (4^{s i n (2) x + c o s (3 x)})

指数の規則を適用する:

a^{b} = e^{b l n (a)}

4^{s i n (2) x + c o s (3 x)} = e^{(s i n (2) x + c o s (3 x)) l n (4)}

= \frac{d}{d x} (e^{(s i n (2) x + c o s (3 x)) l n (4)})

連鎖法則を適用:

e^{(s i n (2) x + c o s (3 x)) l n (4)} \frac{d}{d x} ((sin (2) x + cos (3 x)) ln (4))

= e^{(s i n (2) x + c o s (3 x)) l n (4)} \frac{d}{d x} ((sin (2) x + cos (3 x)) ln (4))

\frac{d}{d x} ((sin (2) x + cos (3 x)) ln (4)) = 2 ln (2) (sin (2) - sin (3 x) \cdot 3)

= e^{(s i n (2) x + c o s (3 x)) l n (4)} \cdot 2 ln (2) (sin (2) - sin (3 x) \cdot 3)

簡素化

e^{(s i n (2) x + c o s (3 x)) l n (4)} \cdot 2 ln (2) (sin (2) - sin (3 x) \cdot 3) : ln (2) \cdot 2^{2 (s i n (2) x + c o s (3 x)) + 1} (sin (2) - 3 sin (3 x))

= ln (2) \cdot 2^{2 (s i n (2) x + c o s (3 x)) + 1} (sin (2) - 3 sin (3 x))