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

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

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

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

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

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

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

Simplify

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

x \to 5 lim (4 x^{2} - 8)

\frac{d}{d x} (x x + 9)

\frac{9}{4} t^{2} e^{5 t}

\int \frac{4 x - 1}{x ^{2} + x - 2} d x

x \to c lim ((8)^{\frac{2}{3}})

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