What is the derivative of 1/(x*ln(x^9)) ?

The derivative of 1/(x*ln(x^9)) is -(ln(x)+1)/(9x^2ln^2(x))

What is the first derivative of 1/(x*ln(x^9)) ?

The first derivative of 1/(x*ln(x^9)) is -(ln(x)+1)/(9x^2ln^2(x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (\frac{1}{x \cdot ln ( x ^{9} )})

- \frac{ln ( x ) + 1}{9 x ^{2} ln ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{1}{x ln ( x ^{9} )})

Apply exponent rule:

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

= \frac{d}{d x} ((x ln (x^{9}))^{- 1})

Apply the chain rule:

- \frac{1}{( x ln ( x ^{9} ) ) ^{2}} \frac{d}{d x} (x ln (x^{9}))

= - \frac{1}{( x ln ( x ^{9} ) ) ^{2}} \frac{d}{d x} (x ln (x^{9}))

\frac{d}{d x} (x ln (x^{9})) = ln (x^{9}) + 9

= - \frac{1}{( x ln ( x ^{9} ) ) ^{2}} (ln (x^{9}) + 9)

Simplify

- \frac{1}{( x ln ( x ^{9} ) ) ^{2}} (ln (x^{9}) + 9) : - \frac{ln ( x ) + 1}{9 x ^{2} ln ^{2} ( x )}

= - \frac{ln ( x ) + 1}{9 x ^{2} ln ^{2} ( x )}

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

\int \frac{8}{x ^{7}} d x

\int 18 x^{17} d x

x + x + x

x^{2} y^{^{'}} + x y = 1

What is the derivative of 1/(x*ln(x^9)) ?
The derivative of 1/(x*ln(x^9)) is -(ln(x)+1)/(9x^2ln^2(x))
What is the first derivative of 1/(x*ln(x^9)) ?
The first derivative of 1/(x*ln(x^9)) is -(ln(x)+1)/(9x^2ln^2(x))