derivative of 3^{-6x+6}

解答

\frac{d}{d x} (3^{- 6 x + 6})

- 2 ln (3) \cdot 3^{- 6 x + 7}

求解步骤

\frac{d}{d x} (3^{- 6 x + 6})

使用指数法则:

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

3^{- 6 x + 6} = e^{(- 6 x + 6) l n (3)}

= \frac{d}{d x} (e^{(- 6 x + 6) l n (3)})

使用连锁律:

e^{(- 6 x + 6) l n (3)} \frac{d}{d x} ((- 6 x + 6) ln (3))

= e^{(- 6 x + 6) l n (3)} \frac{d}{d x} ((- 6 x + 6) ln (3))

\frac{d}{d x} ((- 6 x + 6) ln (3)) = - 6 ln (3)

= e^{(- 6 x + 6) l n (3)} (- 6 ln (3))

化简

e^{(- 6 x + 6) l n (3)} (- 6 ln (3)) : - 2 ln (3) \cdot 3^{- 6 x + 7}

= - 2 ln (3) \cdot 3^{- 6 x + 7}

f^{^{'}} (x) = 3 x^{2} (8 - x^{- 3})

\int_{0}^{1} 5^{2 x} d x

\int_{0}^{1} x^{3} + x d x

n = 0 \sum \infty \frac{x ^{n}}{n ^{2}}

\frac{d ^{2}}{d x ^{2}} (\frac{1}{1 - x})