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

The derivative of ln(x-1+1) is 1/(x-1)

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

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

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

\frac{1}{x - 1}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d x} (1) = 0

= \frac{1}{x - 1} + 0

Simplify

= \frac{1}{x - 1}

x \to 2 lim (x ln (\frac{x}{x - 2}))

f (x) = 4 x^{2} + 5 x + 17

\frac{d}{d x} (4 x - 6)

(- 1.2) (0)

\int (5 x^{3} + 2 x^{2} - 6 x + 3) d x

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