(1/(xln(x)))^'

解答

(\frac{1}{x ln ( x )})^{^{'}}

- \frac{ln ( x ) + 1}{x ^{2} ln ^{2} ( x )}

求解步骤

(\frac{1}{x ln ( x )})^{'}

使用指数法则:

\frac{1}{a} = a^{- 1}

= ((x ln (x))^{- 1})^{^{'}}

使用连锁律:

- \frac{1}{( x ln ( x ) ) ^{2}} (x ln (x))^{'}

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

(x ln (x))^{'} = ln (x) + 1

= - \frac{1}{( x ln ( x ) ) ^{2}} (ln (x) + 1)

化简

- \frac{1}{( x ln ( x ) ) ^{2}} (ln (x) + 1) : - \frac{ln ( x ) + 1}{x ^{2} ln ^{2} ( x )}

= - \frac{ln ( x ) + 1}{x ^{2} ln ^{2} ( x )}

d er i v a t i v e \frac{2 x ^{2} + 6 x + 4}{x}

\frac{d}{d x} (\frac{ln ( 3 x )}{7 x ^{2}})

t an g e n t f (x) = \frac{1}{x}

x \to - 8 lim (\frac{x - 8}{∣ x + 8 ∣})

\int \frac{1}{x ( ln ( x ) ) ^{π}} d x