What is the equation of the normal line of #f(x)=x/lnx# at #x=4 #?

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Answer 1 · 2016-02-17T17:39:22

#y = (ln(4)-1)/(ln(4)^2)(x-4)+4/(ln4)#

First of all, we need a point of intersection. We know the line will intersect #f(x)# at #(4,f(4))# so we will use to obtain the point.

#f(4) = 4/ln(4)#

So we know at least #(4, 4/ln4)# is a point on the line.

Next we need the gradient of the line and this is obtained by taking the derivative of #f(x)#. Using the quotient rule to differentiate we get:

#f'(x) = (ln(x) - 1)/ln(x)^2#

Now find #f'(4)=(ln(4)-1)/ln(4)^2#

Thus we have the gradient of the line.

Now substitute into the formula:

#y-b=m(x-a)# where #m# is the gradient and #(a,b)# is the known point of intersection:

#y-4/(ln4)=(ln(4)-1)/(ln(4)^2)(x-4)#

#y = (ln(4)-1)/(ln(4)^2)(x-4)+4/(ln4)#

We can better visualise this from the graphs of the function:

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