What is the derivative of y=ln(sqrt(2x+1)) ?

The derivative of y=ln(sqrt(2x+1)) is 1/(2x+1)

What is the first derivative of y=ln(sqrt(2x+1)) ?

The first derivative of y=ln(sqrt(2x+1)) is 1/(2x+1)

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derivative of

y = ln (2 x + 1)

\frac{1}{2 x + 1}

Solution steps

\frac{d}{d x} (ln (2 x + 1))

Apply the chain rule:

\frac{1}{2 x + 1} \frac{d}{d x} (2 x + 1)

= \frac{1}{2 x + 1} \frac{d}{d x} (2 x + 1)

\frac{d}{d x} (2 x + 1) = \frac{1}{2 x + 1}

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

Simplify

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

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

\frac{d}{d x} (lo g_{2} (2 x^{2} - 7 x + 1))

\int 4 x e^{9 x} d x

f (x) = x^{4} - 5 x^{2} - x, at x = 1

y = \frac{3}{x}, (5, \frac{3}{5})

f (x) = (2 x + 3) (x^{2} + 2 x)

What is the derivative of y=ln(sqrt(2x+1)) ?
The derivative of y=ln(sqrt(2x+1)) is 1/(2x+1)
What is the first derivative of y=ln(sqrt(2x+1)) ?
The first derivative of y=ln(sqrt(2x+1)) is 1/(2x+1)