What is the tangent of y=ln(x),\at x=2 ?

The tangent of y=ln(x),\at x=2 is y= 1/2 x+ln(2)-1

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tangent of y=ln(x),\at x=2

tangent of

y = ln (x), at x = 2

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

Solution steps

Find the tangent point:

(2, ln (2))

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}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

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

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

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

\int (1 - csc (t) cot (t)) d t

\frac{d}{d x} (a e^{2 x} + b x e^{2 x})

\frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{2}} + 3 x)

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

What is the tangent of y=ln(x),\at x=2 ?
The tangent of y=ln(x),\at x=2 is y= 1/2 x+ln(2)-1