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

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

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tangent of f(x)=ln(x)

tangent of

f (x) = 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

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

Compute the slope of

f (x) = 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

f (x) = 4 x + 1

x \to 0 - lim (x^{2} (sin (\frac{1}{x})))

\frac{d}{d x} (\frac{2 x + 1}{x + 5})

\int (\frac{7}{x} - \frac{6}{x ^{3}}) d x

(x, y) \to (1, - 1) lim ((e^{- x y}) (cos (x + y)))

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