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

Question

Evelyn Agard · Accepted Answer

The normal line has equation #y = 1/2x + 1/2#.

Let's start by finding the corresponding #y#-coordinate that the function #f(x)# passes through. #f(1) = 1/1 - ln(1)/1# #f(1) = 1 - 0/1# #f(1) = 1# Now, let's differentiate using the power rule, the quotient rule and the identity #(lnx)' = 1/x#. #f(x) = x^(-1) - lnx/x# #f'(x) = -1x^(-2) - (1/x xx x - 1 xx lnx)/x^2# #f'(x) = -1/x^2 - (1 - lnx)/x^2# #f'(x) = (-2 + lnx)/x^2# #f'(x) = (lnx - 2)/x^2# We now determine the slope of the tangent by inserting our point into the derivative (since this represents the instantaneous rate of change of the function). #f'(1) = (ln1 - 2)/1^2# #f'(1) = (-2+ 0)/1# #f'(1) = -2# Hence, the slope of the tangent is #-2#. The normal line is always perpendicular to the tangent, so the slope of the normal line is the negative reciprocal of the tangent. #:.# The slope of the normal line is #1/2#. We can now determine the equation of the normal line, because we know the slope #(1)#, and the point of contact #(1, 1)#. #y - y_1 = m(x- x_1)# #y - 1 = 1/2(x - 1)# #y - 1 = 1/2x - 1/2# #y = 1/2x + 1/2# Hopefully this helps!

Emma Aldridge · Answer

undefined The equation of the normal line of f(x)=1/x-lnx/x at x=1 is y = -x + 2.