What is the tangent of ln(sqrt(2x+1)) ?

The tangent of ln(sqrt(2x+1)) is y= 1/(2a_{0)+1}x+1/2 ln(2a_{0}+1)-(a_{0})/(2a_{0)+1}

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tangent of ln(sqrt(2x+1))

tangent of

ln (2 x + 1)

y = \frac{1}{2 a _{0} + 1} x + \frac{1}{2} ln (2 a_{0} + 1) - \frac{a _{0}}{2 a _{0} + 1}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{2} ln (2 a_{0} + 1))

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{2 a _{0} + 1}

Find the line with slope m=

\frac{1}{2 a _{0} + 1}

and passing through

(a_{0}, \frac{1}{2} ln (2 a_{0} + 1)) : y = \frac{1}{2 a _{0} + 1} x + \frac{1}{2} ln (2 a_{0} + 1) - \frac{a _{0}}{2 a _{0} + 1}

y = \frac{1}{2 a _{0} + 1} x + \frac{1}{2} ln (2 a_{0} + 1) - \frac{a _{0}}{2 a _{0} + 1}

x^{2} - 5 x, x + 7

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

\frac{d}{d x} (x - x^{3} sin (x))

\frac{s}{( s ^{2} + 1 ) ( s ^{2} + 2 s + 2 )}

\frac{d y}{d x} = y (x y^{6} + 3)

What is the tangent of ln(sqrt(2x+1)) ?
The tangent of ln(sqrt(2x+1)) is y= 1/(2a_{0)+1}x+1/2 ln(2a_{0}+1)-(a_{0})/(2a_{0)+1}