(\partial)/(\partial b)(10^{-2b})

解答

\frac{\partial}{\partial b} (1 0^{- 2 b})

- ln (10) \cdot 2 5^{- b} \cdot 2^{- 2 b + 1}

求解步骤

\frac{\partial}{\partial b} (1 0^{- 2 b})

化简

1 0^{- 2 b} : 10 0^{- b}

= \frac{\partial}{\partial b} (10 0^{- b})

使用指数法则:

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

10 0^{- b} = e^{(- b) l n (100)}

= \frac{\partial}{\partial b} (e^{(- b) l n (100)})

使用连锁律:

e^{(- b) l n (100)} \frac{\partial}{\partial b} ((- b) ln (100))

= e^{(- b) l n (100)} \frac{\partial}{\partial b} ((- b) ln (100))

\frac{\partial}{\partial b} ((- b) ln (100)) = - 2 ln (10)

= e^{(- b) l n (100)} (- 2 ln (10))

化简

e^{(- b) l n (100)} (- 2 ln (10)) : - ln (10) \cdot 2 5^{- b} \cdot 2^{- 2 b + 1}

= - ln (10) \cdot 2 5^{- b} \cdot 2^{- 2 b + 1}

\int \frac{x ^{2}}{x ^{2} + x - 6} d x

x \to 0 lim (\frac{sin ( - x )}{x})

\frac{d}{d x} (e^{x} + cos (x))

\frac{d}{d x} (10 lo g_{10} (100 x))

\frac{d}{d x} (cos (2 x^{2} - 3))