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

解答

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

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

求解步骤

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

化简

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

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

使用指数法则:

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

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

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

使用连锁律:

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

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

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

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

化简

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

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

\int a rcco tan (x) d x

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a re a x^{2}, x + 2, - 1, 2

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