Question

Calculus I: Implicit differentiation

Answer 1

$2 {x}^{2} + 5x + xy = 3$

$\frac{d}{dx} (2 {x}^{2} + 5x + xy) = 0$

$4x + 5 + y + x \frac{dy}{dx} = 0$

$\frac{dy}{dx} = \frac{ - 4x - y - 5}{x}$

$y'(x) = - 4 - ( \frac{5 + \frac{3 - 2 {x}^{2} - 5x }{x} }{x} )$

$y'(x) = - 4 - \frac{5x + 3 - 2 {x}^{2} - 5x}{ {x}^{2} }$

$y'(x) = - 4 + \frac{2 {x}^{2} - 3}{ {x}^{2} }$

$y'(x) = - 2 - \frac{3}{ {x}^{2} }$

$\boxed{y'(3) = - 2 - \frac{1}{3} = - \frac{7}{3}≈-2.33}$

Thus, The Equation of Tangent at (3,–10) is:

$y + 10 = - \frac{7}{3} (x - 3)$

$- 3y - 30 = 7x - 21$

$\boxed{7x + 3y + 9 = 0}$