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

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

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

tangent of

f (x) = 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

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

Compute the slope of

f (x) = ln (x)

x = 2 : 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{d}{d x} (x^{2} + 2 x + 3)

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

\frac{d}{d x} (\frac{x}{x + 6})

\int_{3}^{6} 21 - \frac{x ^{2}}{2} d x

\int \frac{1}{7 - x ^{2}} d x

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