derivative of 10^{tan(2x})

解

\frac{d}{d x} (1 0^{t a n (2 x)})

ln (10) \cdot 5^{t a n (2 x)} \cdot 2^{t a n (2 x) + 1} sec^{2} (2 x)

解答ステップ

\frac{d}{d x} (1 0^{t a n (2 x)})

指数の規則を適用する:

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

1 0^{t a n (2 x)} = e^{t a n (2 x) l n (10)}

= \frac{d}{d x} (e^{t a n (2 x) l n (10)})

連鎖法則を適用:

e^{t a n (2 x) l n (10)} \frac{d}{d x} (tan (2 x) ln (10))

= e^{t a n (2 x) l n (10)} \frac{d}{d x} (tan (2 x) ln (10))

\frac{d}{d x} (tan (2 x) ln (10)) = ln (10) sec^{2} (2 x) \cdot 2

= e^{t a n (2 x) l n (10)} ln (10) sec^{2} (2 x) \cdot 2

簡素化

e^{t a n (2 x) l n (10)} ln (10) sec^{2} (2 x) \cdot 2 : ln (10) \cdot 5^{t a n (2 x)} \cdot 2^{t a n (2 x) + 1} sec^{2} (2 x)

= ln (10) \cdot 5^{t a n (2 x)} \cdot 2^{t a n (2 x) + 1} sec^{2} (2 x)