What is the equation of the tangent line of #f(x)=lnx# at #x=1#?

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Answer 1 · 2016-07-05T23:41:09

We must first differentiate #f(x)#.

The derivative of #lnx# is #1/x#

The slope of the tangent at a point #x = a# on the function #f(x)# is given by evaluating #f'(a)#.

Hence, the slope of the tangent is #1/1 = 1#.

The function passes through #(1, ln1) = (1, 0)#. Now, we can use point slope form to find the equation of the tangent.

#y - y_1 = m(x - x_1)#

#y - 0 = 1(x - 1)#

#y = x - 1#

Thus, the equation of the tangent is #y = x - 1#.

Hopefully this helps!