# 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 #?

##### 1 Answer

#### 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

#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

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

The normal line passes through the point

#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

#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)#