What is the derivative of 1/((x(ln(x))^p)) ?

The derivative of 1/((x(ln(x))^p)) is -(ln(x)+p)/(x^2ln^{1+p)(x)}

What is the first derivative of 1/((x(ln(x))^p)) ?

The first derivative of 1/((x(ln(x))^p)) is -(ln(x)+p)/(x^2ln^{1+p)(x)}

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derivative of

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

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

Solution steps

\frac{d}{d x} (\frac{1}{x ( ln ( x ) ) ^{p}})

Treat

p

as a constant

Apply exponent rule:

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

= \frac{d}{d x} ((x (ln (x))^{p})^{- 1})

Apply the chain rule:

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

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

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

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

Simplify

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

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

y = 4 x + 1

y = 4 x + 7

(x^{3} + 3 x^{2})^{4}

\frac{( ( x ^{6} ) - 1 )}{( ( x ^{6} ) + 1 )}

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

What is the derivative of 1/((x(ln(x))^p)) ?
The derivative of 1/((x(ln(x))^p)) is -(ln(x)+p)/(x^2ln^{1+p)(x)}
What is the first derivative of 1/((x(ln(x))^p)) ?
The first derivative of 1/((x(ln(x))^p)) is -(ln(x)+p)/(x^2ln^{1+p)(x)}