derivative f(x)= 1/2 b(1/x-1/(x^0))^2

解答

导数

f (x) = \frac{1}{2} b (\frac{1}{x} - \frac{1}{x ^{0}})^{2}

- \frac{b ( 1 - x )}{x ^{3}}

求解步骤

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

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} b \frac{d}{d x} ((\frac{1}{x} - \frac{1}{x ^{0}})^{2})

使用连锁律:

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

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

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

= \frac{1}{2} b \cdot 2 (\frac{1}{x} - \frac{1}{x ^{0}}) (- \frac{1}{x ^{2}})

化简

\frac{1}{2} b \cdot 2 (\frac{1}{x} - \frac{1}{x ^{0}}) (- \frac{1}{x ^{2}}) : - \frac{b ( 1 - x )}{x ^{3}}

= - \frac{b ( 1 - x )}{x ^{3}}

t an g e n t f (x) = x^{3} - 5 x^{2} + 7, at x = 1

\int \frac{x + 5}{x ^{2} + 2 x - 15} d x

\int 0.6 x^{2} d x

x \to + 0 lim (ln (x))

d er i v a t i v e \frac{3 x}{1 - x}