How do you find the equation of the tangent line to the curve # f(x) = ln(x)# at x = 4?

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Answer 1 · 2016-04-02T19:49:03

#y=1/4x+ln4-1#

#f(4)=ln4# so the point #(4,ln4)# lies on the curve and the tangent.

The gradient of the tangent will be the value of the derivative of the function at the point #4#.

#therefore f'(x)=d/dx lnx = 1/x##

#therefore f'(4)=(df)/(dx)|_(x=4)=1/4#

Now since the tangent is a straight line, it has general form #y=mx+c#, where #m# is the gradient and #c# is the y-intercept.

So we may substitute the point #(4,ln4)# as well as the gradient #1/4# into the general equation to solve for #c# :

#therefore ln4=1/4*4+c#

#therefore c=ln4-1#.

Hence the equation of the required tangent line tot eh curve is
#y=1/4x+ln4-1#