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

Dec 23, 2017

Solve for the derivative, plug in $x = 1$ to get the slope of the tangent line, then take the negative reciprocal to get the slope of the normal line, use the point-slope form and finally simplify to get

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

#### Explanation:

First, find the derivative of this function,

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

Using the sum rule, we could split this into two parts,

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

and

$h \left(x\right) = - 2 x$,

so that

$f \left(x\right) = g \left(x\right) + h \left(x\right)$

and

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{dg}}{\mathrm{dx}} + \frac{\mathrm{dh}}{\mathrm{dx}}$.

We could then proceed to find the derivative of each function, starting with $g \left(x\right)$:

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

Using the product rule:

$\mathrm{dg} = {x}^{3} \cdot d \left(\ln \left({x}^{2} + 1\right)\right) + \ln \left({x}^{2} + 1\right) \cdot d \left({x}^{3}\right)$

Let's solve for each differential, from derivatives, starting from $d \left({x}^{3}\right)$:

$\frac{d \left({x}^{3}\right)}{\mathrm{dx}} = 3 {x}^{2} \rightarrow d \left({x}^{3}\right) = 3 {x}^{2} \mathrm{dx}$

Putting that back in:

$\mathrm{dg} = {x}^{3} \cdot d \left(\ln \left({x}^{2} + 1\right)\right) + \ln \left({x}^{2} + 1\right) \cdot 3 {x}^{2} \mathrm{dx}$

Now, how about the other differential? I think solving for that derivative requires breaking it up into smaller functions. Let's set

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

where

${p}_{1} \left(x\right) = {x}^{2}$

${p}_{2} \left(x\right) = x + 1$

${p}_{3} \left(x\right) = \ln \left(x\right)$

then use the chain rule:

$\frac{\mathrm{dp}}{\mathrm{dx}} = \frac{{\mathrm{dp}}_{1}}{\mathrm{dx}} \cdot \frac{{\mathrm{dp}}_{2}}{{\mathrm{dp}}_{1}} \cdot \frac{{\mathrm{dp}}_{3}}{{\mathrm{dp}}_{2}}$

Starting from the derivative of ${p}_{1}$ to $x$:

$\frac{{\mathrm{dp}}_{1}}{\mathrm{dx}} = \frac{d \left({x}^{2}\right)}{\mathrm{dx}} = 2 x$

Then solving for the differential, by "multiplying" by $\mathrm{dx}$:

$\frac{{\mathrm{dp}}_{1}}{\mathrm{dx}} = 2 x \rightarrow {\mathrm{dp}}_{1} = 2 x \mathrm{dx}$

Next, ${p}_{2}$ to ${p}_{1}$:

$\frac{{\mathrm{dp}}_{2}}{{\mathrm{dp}}_{1}} = \frac{d \left({p}_{1} + 1\right)}{{\mathrm{dp}}_{1}} = 1$

$\frac{{\mathrm{dp}}_{2}}{{\mathrm{dp}}_{1}} = 1 \rightarrow {\mathrm{dp}}_{2} = {\mathrm{dp}}_{1}$

Awesome! Evaluate ${\mathrm{dp}}_{1}$:

${\mathrm{dp}}_{2} = 2 x \mathrm{dx}$

Then ${\mathrm{dp}}_{3}$ to ${\mathrm{dp}}_{2}$:

$\frac{{\mathrm{dp}}_{3}}{{\mathrm{dp}}_{2}} = \frac{d \left(\ln \left({p}_{2}\right)\right)}{{\mathrm{dp}}_{2}} = \frac{1}{{p}_{2}}$

$\frac{{\mathrm{dp}}_{3}}{{\mathrm{dp}}_{2}} = \frac{1}{{p}_{2}} \rightarrow {\mathrm{dp}}_{3} = \frac{1}{{p}_{2}} {\mathrm{dp}}_{2}$

${\mathrm{dp}}_{3} = \frac{1}{{p}_{1} + 1} 2 x \mathrm{dx} = \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$

So:

$\frac{{\mathrm{dp}}_{3}}{\mathrm{dx}} = \frac{\mathrm{dp}}{\mathrm{dx}} = \frac{2 x}{{x}^{2} + 1}$

$\frac{\mathrm{dp}}{\mathrm{dx}} = \frac{2 x}{{x}^{2} + 1} \rightarrow \mathrm{dp} = \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$

Substituting $\mathrm{dp}$ back in $\mathrm{dg}$:

$\mathrm{dg} = {x}^{3} \cdot \mathrm{dp} + \ln \left({x}^{2} + 1\right) \cdot 3 {x}^{2} \mathrm{dx}$

$\mathrm{dg} = {x}^{3} \cdot \frac{2 x}{{x}^{2} + 1} \mathrm{dx} + \ln \left({x}^{2} + 1\right) \cdot 3 {x}^{2} \mathrm{dx}$

Simplifying:

$\mathrm{dg} = \frac{2 {x}^{4}}{{x}^{2} + 1} \mathrm{dx} + 3 {x}^{2} \ln \left({x}^{2} + 1\right) \mathrm{dx}$

Dividing by $\mathrm{dx}$ to solve for the derivative:

$\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{2 {x}^{4}}{{x}^{2} + 1} + 3 {x}^{2} \ln \left({x}^{2} + 1\right)$

And substituting this back into $\frac{\mathrm{df}}{\mathrm{dx}}$:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{2 {x}^{4}}{{x}^{2} + 1} + 3 {x}^{2} \ln \left({x}^{2} + 1\right) + \frac{\mathrm{dh}}{\mathrm{dx}}$

Now, on to $h \left(x\right) = - 2 x$:

$\frac{\mathrm{dh}}{\mathrm{dx}} = - 2$

So:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{2 {x}^{4}}{{x}^{2} + 1} + 3 {x}^{2} \ln \left({x}^{2} + 1\right) - 2$

Let's solve for $x = 1$:

$\rightarrow \frac{2 {\left(1\right)}^{4}}{{\left(1\right)}^{2} + 1} + 3 {\left(1\right)}^{2} \ln \left({\left(1\right)}^{2} + 1\right) - 2$

$= 1 + 3 \ln \left(2\right) - 2$

${m}_{t} = 3 \ln \left(2\right) - 1$

That gives us the slope of the tangent line. Taking the negative reciprocal:

${m}_{n} = - {\left({m}_{t}\right)}^{- 1} = - {\left(3 \ln \left(2\right) - 1\right)}^{- 1} = \frac{1}{1 - 3 \ln \left(2\right)}$

Now that we have the slope of the normal line, we just need the $y$-intercept. We can start by using the point-slope form:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

The slope, $m$, in this case is ${m}_{n} = \frac{1}{1 - 3 \ln \left(2\right)}$ :

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

We need a point on the graph. Well, there's our input, $x = 1$, and the output, $f \left(1\right) = {\left(1\right)}^{3} \ln \left({\left(1\right)}^{2} + 1\right) - 2 \left(1\right) = \ln \left(2\right) - 2$, so substitute that for ${x}_{1}$ and ${y}_{1}$ respectively:

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

Finally, isolate $y$:

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

That's the equation for the line normal to $f \left(x\right) = {x}^{3} \ln \left({x}^{2} + 1\right) - 2 x$ at $x = 1$.