What is the tangent of f(x)=ln(x),\at x=6 ?

The tangent of f(x)=ln(x),\at x=6 is y= 1/6 x+ln(6)-1

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tangent of f(x)=ln(x),\at x=6

tangent of

f (x) = ln (x), at x = 6

y = \frac{1}{6} x + ln (6) - 1

Solution steps

Find the tangent point:

(6, ln (6))

Find the slope of

f (x) = ln (x) : \frac{df ( x )}{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}{6}

Find the line with slope m=

\frac{1}{6}

and passing through

(6, ln (6)) : y = \frac{1}{6} x + ln (6) - 1

y = \frac{1}{6} x + ln (6) - 1

\frac{d}{d x} ((\frac{1}{x ^{2}} - \frac{3}{x ^{4}}) (x + 9 x^{3}))

\frac{( 1 - e ^{- s} )}{s ^{2}}

y^{^{''}} + 4 y^{^{'}} + 40 y = 50 cos (8 t)

\int_{- 1}^{1} (1 - y^{2}) - (y^{2} - 1) d y

\int \frac{- 9000}{( 3 x + 50 ) ^{2}} d x

What is the tangent of f(x)=ln(x),\at x=6 ?
The tangent of f(x)=ln(x),\at x=6 is y= 1/6 x+ln(6)-1