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

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

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

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

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\frac{d}{d x} (\frac{1}{x 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 )}

y^{^{'}} = \frac{y}{x} + \frac{x}{y}

y^{^{''}} + y^{^{'}} = sin (20 x)

\frac{d y}{d x} = \frac{1}{10} y cos^{2} (y)

\frac{d}{d x} (\frac{x ^{2} + 32 x}{8})

\int_{0}^{1} 2 π (2 - y) (2 y^{2}) d y

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