Question

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

Answer 1

`y=-5/56(x-2)+8-ln5`

Explanation:

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

`f(2)=(2^3)-ln(2^2+1)=8-ln5`

`f'(x)=3x^2-frac{2x}{x^2+1}`

`f'(2)=3(2)^2-frac{2(2)}{(2)^2+1}=12-(4/5)=56/5`

Slope of the tangent line at `(2, 8-ln5)` is `56/5`, and the slope of the normal line is the negative reciprocal, which is `-5/56`.

Equation of normal line at `(2, 8-ln5)`
`y=-5/56(x-2)+8-ln5`