# 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 ?

May 15, 2017

$y + \frac{1 - 2 \ln \left(2\right)}{2} = \frac{2}{6 \ln \left(2\right) + 3} \left(1 - x\right)$

#### Explanation:

Rewriting:

$f \left(x\right) = {x}^{3} \ln \left({x}^{2} + 1\right) - x {\left({x}^{2} + 1\right)}^{-} 1$

Differentiating with the product rule in both cases:

$f ' \left(x\right) = \left(\frac{d}{\mathrm{dx}} {x}^{3}\right) \ln \left({x}^{2} + 1\right) + {x}^{3} \left(\frac{d}{\mathrm{dx}} \ln \left({x}^{2} + 1\right)\right) - \left(\frac{d}{\mathrm{dx}} x\right) {\left({x}^{2} + 1\right)}^{-} 1 - x \left(\frac{d}{\mathrm{dx}} {\left({x}^{2} + 1\right)}^{-} 1\right)$

The chain rule will apply twice:

$f ' \left(x\right) = 3 {x}^{2} \ln \left({x}^{2} + 1\right) + {x}^{3} \left(\frac{1}{{x}^{2} + 1}\right) \left(\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)\right) - {\left({x}^{2} + 1\right)}^{-} 1 - x \left(- {\left({x}^{2} + 1\right)}^{-} 2\right) \left(\frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)\right)$

$f ' \left(x\right) = 3 {x}^{2} \ln \left({x}^{2} + 1\right) + {x}^{3} / \left({x}^{2} + 1\right) \left(3 {x}^{2}\right) - \frac{1}{{x}^{2} + 1} + \frac{x}{{x}^{2} + 1} ^ 2 \left(2 x\right)$

$f ' \left(x\right) = 3 {x}^{2} \ln \left({x}^{2} + 1\right) + \frac{3 {x}^{5} - 1}{{x}^{2} + 1} + \frac{2 {x}^{2}}{{x}^{2} + 1} ^ 2$

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

$f ' \left(1\right) = 3 {\left(1\right)}^{2} \ln \left({1}^{2} + 1\right) + \frac{3 {\left(1\right)}^{5} - 1}{{1}^{2} + 1} + \frac{2 {\left(1\right)}^{2}}{{1}^{2} + 1} ^ 2$

$f ' \left(1\right) = 3 \ln \left(2\right) + \frac{2}{2} + \frac{2}{4}$

$f ' \left(1\right) = \frac{6 \ln \left(2\right) + 3}{2}$

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

$m = \frac{- 1}{f ' \left(1\right)} = \frac{- 2}{6 \ln \left(2\right) + 3}$

The normal line passes through the point $\left(1 , f \left(1\right)\right)$. Note that:

$f \left(1\right) = {\left(1\right)}^{3} \ln \left({1}^{2} + 1\right) - \frac{1}{{1}^{2} + 1}$

$f \left(1\right) = \ln \left(2\right) - \frac{1}{2}$

$f \left(1\right) = \frac{2 \ln \left(2\right) - 1}{2}$

The line that passes through the point $\left(1 , \frac{2 \ln \left(2\right) - 1}{2}\right)$ with slope $\frac{- 2}{6 \ln \left(2\right) + 3}$ is given by:

$y - \frac{2 \ln \left(2\right) - 1}{2} = \frac{- 2}{6 \ln \left(2\right) + 3} \left(x - 1\right)$

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

$y + \frac{1 - 2 \ln \left(2\right)}{2} = \frac{2}{6 \ln \left(2\right) + 3} \left(1 - x\right)$