What is the equation of the normal line of f(x)=1x−ln⁡xx at x=1?

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Cesar Mcfarland · Accepted Answer

Let&#039;s start by finding the corresponding y-coordinate that the function f(x) passes through.f(1)=11−ln⁡(1)1f(1)=1−01f(1)=1Now, let&#039;s differentiate using the power rule, the quotient rule and the identity(ln⁡x)′=1xf(x)=x−1−ln⁡xxf(x)=x−1−ln⁡xxf′(x)=−1x−2−1x×x−1×ln⁡xx2f′(x)=−1x2−1−ln⁡xx2f′(x)=−2+ln⁡xx2f′(x)=ln⁡x−2x2We 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)=ln⁡1−212f′(1)=−2+01f&#039;(1)=-2Hence, 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 12.We can now determine the equation of the normal line, because we know the slope(1), and the point of contact (1,1).y−y1=m(x−x1)y−1=12(x−1)y−1=12x−12y=12x+12

What is the equation of the normal line

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