What is the tangent of f(x)=ln(sqrt(2x+1)),\at x=0 ?

The tangent of f(x)=ln(sqrt(2x+1)),\at x=0 is y=x

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tangent of f(x)=ln(sqrt(2x+1)),\at x=0

tangent of

f (x) = ln (2 x + 1), at x = 0

y = x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = ln (2 x + 1) : \frac{df ( x )}{d x} = \frac{1}{2 x + 1}

Compute the slope of

f (x) = ln (2 x + 1)

x = 0 : m = 1

Find the line with slope m=

1

and passing through

(0, 0) : y = x

y = x

u (x - 5)

f (x) = \frac{3}{x}, at x = 4

\frac{d}{d x} (\frac{4 x ^{3}}{9})

\int_{- 4}^{- 2} (x^{2} - 4) d x

5 x^{2} + x

What is the tangent of f(x)=ln(sqrt(2x+1)),\at x=0 ?
The tangent of f(x)=ln(sqrt(2x+1)),\at x=0 is y=x