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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

x y^{^{'}} - y = \frac{5}{4} x ln (x)

(6 - x) x + 15

5 e^{- 3 t}

\frac{d}{d x} (A (e^{- x} - e^{- x} x))

y = 8 x arccos (x - 1), at x = 1

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