Question

What is the equation of the line that is normal to `f(x)= x^3 ln(x^2+1)-x/(x^2+1)`at `x= 1`?

Answer 1

`y+(1-2ln(2))/2=2/(6ln(2)+3)(1-x)`

Explanation:

Rewriting:

`f(x)=x^3ln(x^2+1)-x(x^2+1)^-1`

Differentiating with the product rule in both cases:

`f'(x)=(d/dxx^3)ln(x^2+1)+x^3(d/dxln(x^2+1))-(d/dxx)(x^2+1)^-1-x(d/dx(x^2+1)^-1)`

The chain rule will apply twice:

`f'(x)=3x^2ln(x^2+1)+x^3(1/(x^2+1))(d/dx(x^2+1))-(x^2+1)^-1-x(-(x^2+1)^-2)(d/dx(x^2+1))`

`f'(x)=3x^2ln(x^2+1)+x^3/(x^2+1)(3x^2)-1/(x^2+1)+x/(x^2+1)^2(2x)`

`f'(x)=3x^2ln(x^2+1)+(3x^5-1)/(x^2+1)+(2x^2)/(x^2+1)^2`

The slope of the tangent line at `x=1` is:

`f'(1)=3(1)^2ln(1^2+1)+(3(1)^5-1)/(1^2+1)+(2(1)^2)/(1^2+1)^2`

`f'(1)=3ln(2)+2/2+2/4`

`f'(1)=(6ln(2)+3)/2`

So the slope of the normal line, which is perpendicular to the tangent line, is the opposite reciprocal of `f'(1)`. This means the normal line has a slope of:

`m=(-1)/(f'(1))=(-2)/(6ln(2)+3)`

The normal line passes through the point `(1,f(1))`. Note that:

`f(1)=(1)^3ln(1^2+1)-1/(1^2+1)`

`f(1)=ln(2)-1/2`

`f(1)=(2ln(2)-1)/2`

The line that passes through the point `(1,(2ln(2)-1)/2)` with slope `(-2)/(6ln(2)+3)` is given by:

`y-(2ln(2)-1)/2=(-2)/(6ln(2)+3)(x-1)`

Which, if we like, can be "simplified" as:

`y+(1-2ln(2))/2=2/(6ln(2)+3)(1-x)`