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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int x^{2} 3 + 7 x^{2} d x

\frac{d}{d x} (ln (y - x^{2}))

\int x \cdot e^{x^{2} + 1} d x

(5 x + y e^{\frac{y}{x}}) d x - (x e^{\frac{y}{x}}) d y = 0, y (1) = 3

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

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