derivative 15000+2500^t

解答

导数

15000 + 250 0^{t}

ln (50) \cdot 62 5^{t} \cdot 2^{2 t + 1}

求解步骤

\frac{d}{d t} (15000 + 250 0^{t})

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d t} (15000) + \frac{d}{d t} (250 0^{t})

\frac{d}{d t} (15000) = 0

\frac{d}{d t} (250 0^{t}) = ln (50) \cdot 62 5^{t} \cdot 2^{2 t + 1}

= 0 + ln (50) \cdot 62 5^{t} \cdot 2^{2 t + 1}

化简

= ln (50) \cdot 62 5^{t} \cdot 2^{2 t + 1}

\frac{\partial}{\partial x} (\frac{2}{1 + x ^{3}})

\int x ln^{2} (d) x d x

\frac{d}{d x} (arccot (x^{2}))

\frac{d}{d x} (3 e^{- 5 x})

d er i v a t i v e f (x) = x^{3} - 12 x^{2} + 45 x