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

The (sqrt(2x+1))^' is 1/(sqrt(2x+1))

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

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

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

\frac{1}{2 x + 1}

Solution steps

(2 x + 1)^{'}

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d}{d x} (2 x^{4} + x^{3} + 3 x^{2})

f (x) = \frac{1}{x - 3}, at x = 8

\int t^{2} sin (t) d t

f (t) = cos^{2} (\frac{π}{2} - t)

\frac{d}{d x} (2 x^{2} - 5 x + 2)