What is the tangent of f(x)=ln(x),\at x=1 ?

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

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tangent of f(x)=ln(x),\at x=1

tangent of

f (x) = ln (x), at x = 1

y = x - 1

Solution steps

Find the tangent point:

(1, 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 = 1 : m = 1

Find the line with slope m=

1

and passing through

(1, 0) : y = x - 1

y = x - 1

(43, 4)

x

y = \frac{2}{3} x

f (x) = 5 x^{2} (x + 47)

y = csc (x)

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