((x^4)/8+1/(4x^2))^'

解答

(\frac{x ^{4}}{8} + \frac{1}{4 x ^{2}})^{^{'}}

\frac{x ^{3}}{2} - \frac{1}{2 x ^{3}}

求解步骤

(\frac{x ^{4}}{8} + \frac{1}{4 x ^{2}})^{'}

使用微分加减法定则:

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

= (\frac{x ^{4}}{8})^{^{'}} + (\frac{1}{4 x ^{2}})^{^{'}}

(\frac{x ^{4}}{8})^{'} = \frac{x ^{3}}{2}

(\frac{1}{4 x ^{2}})^{'} = - \frac{1}{2 x ^{3}}

= \frac{x ^{3}}{2} - \frac{1}{2 x ^{3}}

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

\int e^{- 3 y} d y

x \to 6 lim (x + 6)

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

t an g e n t f (x) = x^{2} - x + 1