# Question #0ed82

Nov 19, 2016

$y ' \left(5\right) = - \frac{16}{5}$

#### Explanation:

Making use of implicit differentiation and the product rule, we have

$\frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 3 x + x y\right) = \frac{d}{\mathrm{dx}} 5$

$\implies \left(\frac{d}{\mathrm{dx}} 3 {x}^{2}\right) + \left(\frac{d}{\mathrm{dx}} 3 x\right) + \left(\frac{d}{\mathrm{dx}} x y\right) = 0$

$\implies 6 x + 3 + \left[x \left(\frac{d}{\mathrm{dx}} y\right) + y \left(\frac{d}{\mathrm{dx}} x\right)\right] = 0$

$\implies 6 x + 3 + x \frac{\mathrm{dy}}{\mathrm{dx}} + y = 0$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{6 x + y + 3}{x}$

Evaluated at $x = 5$, and noting that $x = 5$ implies $y = y \left(5\right) = - 17$, we have

$y ' \left(5\right) = \frac{\mathrm{dy}}{\mathrm{dx}} {|}_{x = 5}$

$= - \frac{6 \left(5\right) + \left(- 17\right) + 3}{5}$

$= - \frac{30 - 17 + 3}{5}$

$= - \frac{16}{5}$