What is the derivative of (ln(x-1)/((x-1)^3)) ?

The derivative of (ln(x-1)/((x-1)^3)) is (1-3ln(x-1))/((x-1)^4)

What is the first derivative of (ln(x-1)/((x-1)^3)) ?

The first derivative of (ln(x-1)/((x-1)^3)) is (1-3ln(x-1))/((x-1)^4)

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\frac{d}{d x} (\frac{ln ( x - 1 )}{( x - 1 ) ^{3}})

\frac{1 - 3 ln ( x - 1 )}{( x - 1 ) ^{4}}

Solution steps

\frac{d}{d x} (\frac{ln ( x - 1 )}{( x - 1 ) ^{3}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( ln ( x - 1 ) ) ( x - 1 ) ^{3} - \frac{d}{d x} ( ( x - 1 ) ^{3} ) ln ( x - 1 )}{( ( x - 1 ) ^{3} ) ^{2}}

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

\frac{d}{d x} ((x - 1)^{3}) = 3 (x - 1)^{2}

= \frac{\frac{1}{x - 1} ( x - 1 ) ^{3} - 3 ( x - 1 ) ^{2} ln ( x - 1 )}{( ( x - 1 ) ^{3} ) ^{2}}

Simplify

\frac{\frac{1}{x - 1} ( x - 1 ) ^{3} - 3 ( x - 1 ) ^{2} ln ( x - 1 )}{( ( x - 1 ) ^{3} ) ^{2}} : \frac{1 - 3 ln ( x - 1 )}{( x - 1 ) ^{4}}

= \frac{1 - 3 ln ( x - 1 )}{( x - 1 ) ^{4}}

f (x) = cos (2) (x)

\int \frac{1}{25 - 9 x ^{2}} d x

y = ln (\frac{5 + e ^{x}}{5 - e ^{x}})

\int x 3 x - 4 d x

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

What is the derivative of (ln(x-1)/((x-1)^3)) ?
The derivative of (ln(x-1)/((x-1)^3)) is (1-3ln(x-1))/((x-1)^4)
What is the first derivative of (ln(x-1)/((x-1)^3)) ?
The first derivative of (ln(x-1)/((x-1)^3)) is (1-3ln(x-1))/((x-1)^4)