What is the tangent of ln(x) ?

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

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tangent of ln(x)

tangent of

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}

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}{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

\frac{d}{d x} ((x^{2} + y^{2} + z^{2})^{\frac{1}{2}})

\frac{d}{d x} (6 ln (sec (x) + tan (x)))

cos^{3} (x)

\frac{1}{s ^{2} ( s ^{2} + 9 )}

y^{^{'}} - \frac{6}{x} y = \frac{y ^{4}}{x ^{15}}

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