What is the first (ln(2x+1))^' ?

The first (ln(2x+1))^' is 2/(2x+1)

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(ln(2x+1))^'

(ln (2 x + 1))^{^{'}}

\frac{2}{2 x + 1}

Solution steps

(ln (2 x + 1))^{'}

Apply the chain rule:

\frac{1}{2 x + 1} (2 x + 1)^{'}

= \frac{1}{2 x + 1} (2 x + 1)^{'}

(2 x + 1)^{'} = 2

= \frac{1}{2 x + 1} \cdot 2

Simplify

\frac{1}{2 x + 1} \cdot 2 : \frac{2}{2 x + 1}

= \frac{2}{2 x + 1}

(2 x y) d x - (1 + y) d y = 0

y^{^{''}} - 8 y^{^{'}} - 9 y = 21 e^{2 t}

x \to 5 - lim (\frac{7}{x - 5})

d er i v a t i v e (x - 2) (x + 1) (x + 6)

t an g e n t y = 4 + 8 x^{2}, (0, 4)