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

Mar 5, 2017

$y = - \frac{5}{56} \left(x - 2\right) + 8 - \ln 5$

#### Explanation:

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

$f \left(2\right) = \left({2}^{3}\right) - \ln \left({2}^{2} + 1\right) = 8 - \ln 5$

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

$f ' \left(2\right) = 3 {\left(2\right)}^{2} - \frac{2 \left(2\right)}{{\left(2\right)}^{2} + 1} = 12 - \left(\frac{4}{5}\right) = \frac{56}{5}$

Slope of the tangent line at $\left(2 , 8 - \ln 5\right)$ is $\frac{56}{5}$, and the slope of the normal line is the negative reciprocal, which is $- \frac{5}{56}$.

Equation of normal line at $\left(2 , 8 - \ln 5\right)$
$y = - \frac{5}{56} \left(x - 2\right) + 8 - \ln 5$