# Question #15785

May 17, 2017

$- \frac{12}{5}$

#### Explanation:

To make the function easier to derive we will isolate $y$.
$5 \ln y = 12 - 3 {x}^{2}$
$\ln y = \frac{12 - 3 {x}^{2}}{5}$
$y = {e}^{\frac{12 - 3 {x}^{2}}{5}}$

The we find the tangent line, which will be the derivative
$y ' = {e}^{\frac{12 - 3 {x}^{2}}{5}} \left(- \frac{6 x}{5}\right)$

We then plug in $x = 2$ to get
$y ' \left(2\right) = {e}^{\frac{12 - 3 {\left(2\right)}^{2}}{5}} \left(- \frac{6 \left(2\right)}{5}\right)$
$y ' \left(2\right) = {e}^{\frac{0}{5}} \left(- \frac{12}{5}\right)$
$y ' \left(2\right) = 1 \left(- \frac{12}{5}\right)$
$y ' \left(2\right) = - \frac{12}{5}$