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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int_{0}^{1} \frac{1}{x ^{\frac{1}{3}}} d x

2 x^{2}, 8 x - 2 x^{2}

(2 x - 1) d x + (5 y + 3) d y = 0

\int 1 + \frac{16}{x ^{\frac{2}{3}}} d x

\int_{4}^{2} x d x

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