What is the first (ln(x+1))^' ?

The first (ln(x+1))^' is 1/(x+1)

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(ln(x+1))^'

(ln (x + 1))^{^{'}}

\frac{1}{x + 1}

Solution steps

(ln (x + 1))^{'}

Apply the chain rule:

\frac{1}{x + 1} (x + 1)^{'}

= \frac{1}{x + 1} (x + 1)^{'}

(x + 1)^{'} = 1

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

Simplify

= \frac{1}{x + 1}

x^{2} + 1, 0, 1

x \to \infty lim (1.0 1^{x})

(x^{2} - x + 1) y^{^{''}} - (x^{2} + x) y^{^{'}} + (x + 1) y = 0

y = 2187 x, y = 81 x^{2}

6 sin (x)