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

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

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Answer 1 · 2017-12-23T08:32:34

Solve for the derivative, plug in $x = 1$ $x = 1$ 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

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

Explanation:

First, find the derivative of this function,

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

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

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

and

$h (x) = - 2 x$ $h(x) = -2x$ ,

so that

$f (x) = g (x) + h (x)$ $f(x) = g(x) + h(x)$

and

$\frac{d f}{d x} = \frac{d g}{d x} + \frac{d h}{d x}$ $(df)/(dx) = (dg)/(dx) + (dh)/(dx)$ .

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

$\frac{d g}{d x} = \frac{d}{d x} (x^{3} ln (x^{2} + 1))$ $(dg)/(dx) = d/dx (x^3 ln(x^2 + 1))$

Using the product rule:

$d g = x^{3} \cdot d (ln (x^{2} + 1)) + ln (x^{2} + 1) \cdot d (x^{3})$ $dg = x^3 * d(ln(x^2 + 1)) + ln(x^2 + 1) * d(x^3)$

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

$\frac{d (x^{3})}{d x} = 3 x^{2} \to d (x^{3}) = 3 x^{2} d x$ $(d(x^3))/(dx) = 3x^2 rarr d(x^3) = 3x^2 dx$

Putting that back in:

$d g = x^{3} \cdot d (ln (x^{2} + 1)) + ln (x^{2} + 1) \cdot 3 x^{2} d x$ $dg = x^3 * d(ln(x^2 + 1)) + ln(x^2 + 1) * 3x^2 dx$

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

$p (x) = ln (x^{2} + 1) = p_{3} (p_{2} (p_{1} (x)))$ $p(x) = ln(x^2 + 1) = p_3(p_2(p_1(x)))$

where

$p_{1} (x) = x^{2}$ $p_1(x) = x^2$

$p_{2} (x) = x + 1$ $p_2(x) = x + 1$

$p_{3} (x) = ln (x)$ $p_3(x) = ln(x)$

then use the chain rule:

$\frac{d p}{d x} = \frac{d p_{1}}{d x} \cdot \frac{d p_{2}}{d p_{1}} \cdot \frac{d p_{3}}{d p_{2}}$ $(dp)/(dx) = (dp_1)/(dx) * (dp_2)/(dp_1) * (dp_3)/(dp_2)$

Starting from the derivative of $p_{1}$ $p_1$ to $x$ $x$ :

$\frac{d p_{1}}{d x} = \frac{d (x^{2})}{d x} = 2 x$ $(dp_1)/(dx) = (d(x^2))/(dx) = 2x$

Then solving for the differential, by "multiplying" by $d x$ $dx$ :

$\frac{d p_{1}}{d x} = 2 x \to d p_{1} = 2 x d x$ $(dp_1)/(dx) = 2x rarr dp_1 = 2x dx$

Next, $p_{2}$ $p_2$ to $p_{1}$ $p_1$ :

$\frac{d p_{2}}{d p_{1}} = \frac{d (p_{1} + 1)}{d p_{1}} = 1$ $(dp_2)/(dp_1) = (d(p_1 + 1))/(dp_1) = 1$

$\frac{d p_{2}}{d p_{1}} = 1 \to d p_{2} = d p_{1}$ $(dp_2)/(dp_1) = 1 rarr dp_2 = dp_1$

Awesome! Evaluate $d p_{1}$ $dp_1$ :

$d p_{2} = 2 x d x$ $dp_2 = 2x dx$

Then $d p_{3}$ $dp_3$ to $d p_{2}$ $dp_2$ :

$\frac{d p_{3}}{d p_{2}} = \frac{d (ln (p_{2}))}{d p_{2}} = \frac{1}{p_{2}}$ $(dp_3)/(dp_2) = (d(ln(p_2)))/(dp_2) = 1/(p_2)$

$\frac{d p_{3}}{d p_{2}} = \frac{1}{p_{2}} \to d p_{3} = \frac{1}{p_{2}} d p_{2}$ $(dp_3)/(dp_2) = 1/(p_2) rarr dp_3 = 1/(p_2) dp_2$

$d p_{3} = \frac{1}{p_{1} + 1} 2 x d x = \frac{2 x}{x^{2} + 1} d x$ $dp_3 = 1/(p_1 + 1) 2x dx = (2x)/(x^2 + 1) dx$

So:

$\frac{d p_{3}}{d x} = \frac{d p}{d x} = \frac{2 x}{x^{2} + 1}$ $(dp_3)/(dx) = (dp)/(dx) = (2x)/(x^2 + 1)$

$\frac{d p}{d x} = \frac{2 x}{x^{2} + 1} \to d p = \frac{2 x}{x^{2} + 1} d x$ $(dp)/(dx) = (2x)/(x^2 + 1) rarr dp = (2x)/(x^2 + 1) dx$

Substituting $d p$ $dp$ back in $d g$ $dg$ :

$d g = x^{3} \cdot d p + ln (x^{2} + 1) \cdot 3 x^{2} d x$ $dg = x^3 * dp + ln(x^2 + 1) * 3x^2 dx$

$d g = x^{3} \cdot \frac{2 x}{x^{2} + 1} d x + ln (x^{2} + 1) \cdot 3 x^{2} d x$ $dg = x^3 * (2x)/(x^2 + 1) dx + ln(x^2 + 1) * 3x^2 dx$

Simplifying:

$d g = \frac{2 x^{4}}{x^{2} + 1} d x + 3 x^{2} ln (x^{2} + 1) d x$ $dg = (2x^4)/(x^2 + 1) dx + 3x^2 ln(x^2 + 1) dx$

Dividing by $d x$ $dx$ to solve for the derivative:

$\frac{d g}{d x} = \frac{2 x^{4}}{x^{2} + 1} + 3 x^{2} ln (x^{2} + 1)$ $(dg)/(dx) = (2x^4)/(x^2 + 1) + 3x^2 ln(x^2 + 1)$

And substituting this back into $\frac{d f}{d x}$ $(df)/(dx)$ :

$\frac{d f}{d x} = \frac{2 x^{4}}{x^{2} + 1} + 3 x^{2} ln (x^{2} + 1) + \frac{d h}{d x}$ $(df)/(dx) = (2x^4)/(x^2 + 1) + 3x^2 ln(x^2 + 1) + (dh)/(dx)$

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

$\frac{d h}{d x} = - 2$ $(dh)/(dx) = -2$

So:

$\frac{d f}{d x} = \frac{2 x^{4}}{x^{2} + 1} + 3 x^{2} ln (x^{2} + 1) - 2$ $(df)/(dx) = (2x^4)/(x^2 + 1) + 3x^2 ln(x^2 + 1) - 2$

Let's solve for $x = 1$ $x = 1$ :

$\to \frac{2 {(1)}^{4}}{{(1)}^{2} + 1} + 3 {(1)}^{2} ln ({(1)}^{2} + 1) - 2$ $rarr (2(1)^4)/((1)^2 + 1) + 3(1)^2 ln((1)^2 + 1) - 2$

$= 1 + 3 ln (2) - 2$ $= 1 + 3 ln(2) - 2$

$m_{t} = 3 ln (2) - 1$ $m_t = 3 ln(2) - 1$

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

$m_{n} = - {(m_{t})}_{t}^{- 1} = - {(3 ln (2) - 1)}^{- 1} = \frac{1}{1 - 3 ln (2)}$ $m_n = -(m_t)^(-1) = -(3 ln(2) - 1)^(-1) = 1/(1 - 3 ln(2))$

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

$y - y_{1} = m (x - x_{1})$ $y - y_1 = m(x - x_1)$

The slope, $m$ $m$ , in this case is $m_{n} = \frac{1}{1 - 3 ln (2)}$ $m_n = 1/(1 - 3 ln(2))$ :

$y - y_{1} = \frac{x - x_{1}}{1 - 3 ln (2)}$ $y - y_1 = (x - x_1)/(1 - 3 ln(2))$

We need a point on the graph. Well, there's our input, $x = 1$ $x = 1$ , and the output, $f (1) = {(1)}^{3} ln ({(1)}^{2} + 1) - 2 (1) = ln (2) - 2$ $f(1) = (1)^3 ln((1)^2 + 1) - 2(1) = ln(2) - 2$ , so substitute that for $x_{1}$ $x_1$ and $y_{1}$ $y_1$ respectively:

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

Finally, isolate $y$ $y$ :

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

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

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

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