# What is the slope of the line normal to the tangent line of f(x) = xe^(x-3)+x^3  at  x= 2 ?

Feb 29, 2016

$- \frac{e}{3 + 12 e}$

#### Explanation:

Before anything else, find the function's derivative. To do this, we will need the product rule for the $x {e}^{x - 3}$ term. For ${x}^{3}$, just use the power rule.

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

The simpler derivative here is $\frac{d}{\mathrm{dx}} \left(x\right) = 1$. The more difficult is $\frac{d}{\mathrm{dx}} \left({e}^{x - 3}\right)$, for which we will need to use chain rule.

Since $\frac{d}{\mathrm{dx}} \left({e}^{x}\right) = {e}^{x}$, we can say that $\frac{d}{\mathrm{dx}} \left({e}^{g \left(x\right)}\right) = {e}^{g \left(x\right)} \cdot g ' \left(x\right)$.

Thus, $\frac{d}{\mathrm{dx}} \left({e}^{x - 3}\right) = {e}^{x - 3} \frac{d}{\mathrm{dx}} \left(x - 3\right) = {e}^{x - 3} \left(1\right) = {e}^{x - 3}$.

Plugging this back into our $f ' \left(x\right)$ equation, we see that

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

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

Now, we must find the slope of the tangent line, which is equal to the value of the derivative at $x = 2$. From here, we will be able to find the slope of the normal line, since the two are perpendicular.

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

This, however, is the slope of the tangent line. Since perpendicular lines have opposite reciprocal slopes, the slope of the normal line is

$- \frac{1}{\frac{3}{e} + 12} = - \frac{1}{\frac{3 + 12 e}{e}} = - \frac{e}{3 + 12 e}$