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

Nov 8, 2016

$y - \ln \left(2 + 2 {e}^{6}\right) = - \frac{2 + 2 {e}^{6}}{1 + 7 {e}^{6}} \left(x - 2\right)$

#### Explanation:

We can rewrite the equation as follows (and consequently avoid needing to use the product rule):

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

We can differentiate (using the chain rule):
$\therefore f ' \left(x\right) = \frac{1}{x} + \frac{1}{1 + {e}^{3 x}} \cdot \left(3 {e}^{3 x}\right)$
$\therefore f ' \left(x\right) = \frac{1}{x} + \frac{3 {e}^{3 x}}{1 + {e}^{3 x}}$
$\therefore f ' \left(x\right) = \frac{\left(1 + {e}^{3 x}\right) + \left(3 x {e}^{3 x}\right)}{x \left(1 + {e}^{3 x}\right)}$

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

When $x = 2 \implies f \left(2\right) = \ln \left(2 + 2 {e}^{6}\right)$
And, $f ' \left(2\right) = \frac{6 {e}^{6} + {e}^{6} + 1}{2 + 2 {e}^{6}} = \frac{1 + 7 {e}^{6}}{2 + 2 {e}^{6}}$

This is the gradient of the tangent when $x = 2$. As the normal and tangent are perpendicular the product of their gradients is $- 1$, so the normal passes through $\left(2 , \ln \left(2 + 2 {e}^{6}\right)\right)$ and it has gradient $m = - \frac{2 + 2 {e}^{6}}{1 + 7 {e}^{6}}$

So using $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the required equation is;

$y - \ln \left(2 + 2 {e}^{6}\right) = - \frac{2 + 2 {e}^{6}}{1 + 7 {e}^{6}} \left(x - 2\right)$