What is the derivative of y= 1/(x(8+ln(x))) ?

The derivative of y= 1/(x(8+ln(x))) is -(ln(x)+9)/(x^2(8+ln(x))^2)

What is the first derivative of y= 1/(x(8+ln(x))) ?

The first derivative of y= 1/(x(8+ln(x))) is -(ln(x)+9)/(x^2(8+ln(x))^2)

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derivative of

y = \frac{1}{x ( 8 + ln ( x ) )}

- \frac{ln ( x ) + 9}{x ^{2} ( 8 + ln ( x ) ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{x ( 8 + ln ( x ) )})

Apply exponent rule:

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

= \frac{d}{d x} ((x (8 + ln (x)))^{- 1})

Apply the chain rule:

- \frac{1}{( x ( 8 + ln ( x ) ) ) ^{2}} \frac{d}{d x} (x (8 + ln (x)))

= - \frac{1}{( x ( 8 + ln ( x ) ) ) ^{2}} \frac{d}{d x} (x (8 + ln (x)))

\frac{d}{d x} (x (8 + ln (x))) = ln (x) + 9

= - \frac{1}{( x ( 8 + ln ( x ) ) ) ^{2}} (ln (x) + 9)

Simplify

- \frac{1}{( x ( 8 + ln ( x ) ) ) ^{2}} (ln (x) + 9) : - \frac{ln ( x ) + 9}{x ^{2} ( 8 + ln ( x ) ) ^{2}}

= - \frac{ln ( x ) + 9}{x ^{2} ( 8 + ln ( x ) ) ^{2}}

x y^{^{'}} + y = - 3 x y^{2}

\int e^{- 3 x} cos (2 x) d x

(5 x + 2) + (5 y - 5) y^{^{'}} = 0

\frac{4 x}{x ^{2} + 1} (1.1)

\int (3 x^{8}) d x

What is the derivative of y= 1/(x(8+ln(x))) ?
The derivative of y= 1/(x(8+ln(x))) is -(ln(x)+9)/(x^2(8+ln(x))^2)
What is the first derivative of y= 1/(x(8+ln(x))) ?
The first derivative of y= 1/(x(8+ln(x))) is -(ln(x)+9)/(x^2(8+ln(x))^2)