What is the (1/(xln(x)))^' ?

The (1/(xln(x)))^' is -(ln(x)+1)/(x^2ln^2(x))

What is the first (1/(xln(x)))^' ?

The first (1/(xln(x)))^' is -(ln(x)+1)/(x^2ln^2(x))

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

- \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)

Simplify

- \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 )}

\frac{2 x ^{2} + 6 x + 4}{x}

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

f (x) = \frac{1}{x}

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

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

What is the (1/(xln(x)))^' ?
The (1/(xln(x)))^' is -(ln(x)+1)/(x^2ln^2(x))
What is the first (1/(xln(x)))^' ?
The first (1/(xln(x)))^' is -(ln(x)+1)/(x^2ln^2(x))