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Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
x^2+ 2xy - y^2 + x = 20, (3, 4) (hyperbola)

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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