What is the tangent of y=ln(x) ?

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

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tangent of y=ln(x)

tangent of

y = ln (x)

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, ln (a_{0}))

Find the slope of

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

Compute the slope of

y = ln (x)

x = a_{0} : m = \frac{1}{a _{0}}

Find the line with slope m=

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

and passing through

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

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

(- 6, 0), (0, 1)

f (x) = a x^{2} + b x + c

f (x) = x + 2

(5, 5)

y = 2

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