Question

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

Answer 1

Differentiating both sides of

$x^2 + 2xy - y^2 + x = 20$

with respect to $x$ yields

$2x + 2y + 2x \dfrac{dy}{dx} - 2y \dfrac{dy}{dx} + 1 = 0 \\\\ \implies (2x-2y) \dfrac{dy}{dx} = -1 - 2x - 2y \\\\ \implies \dfrac{dy}{dx} = \dfrac{1 + 2x + 2y}{2(y-x)}$

At the point (3, 4) (so $x=3$ and $y=4$), the tangent line has slope

$\dfrac{dy}{dx} = \dfrac{1 + 2\times3 + 2\times4}{2(4-3)} = \dfrac{15}2$

Then the tangent line to (3, 4) has equation

$y - 4 = \dfrac{15}2 (x - 3) \implies \boxed{y = \dfrac{15}2 x - \dfrac{37}2}$