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Sagot :
Differentiating both sides of
[tex]x^2 + 2xy - y^2 + x = 20[/tex]
with respect to [tex]x[/tex] yields
[tex]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)}[/tex]
At the point (3, 4) (so [tex]x=3[/tex] and [tex]y=4[/tex]), the tangent line has slope
[tex]\dfrac{dy}{dx} = \dfrac{1 + 2\times3 + 2\times4}{2(4-3)} = \dfrac{15}2[/tex]
Then the tangent line to (3, 4) has equation
[tex]y - 4 = \dfrac{15}2 (x - 3) \implies \boxed{y = \dfrac{15}2 x - \dfrac{37}2}[/tex]
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