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Jet001 [13]
1 year ago
9

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

Mathematics
1 answer:
andrew11 [14]1 year ago
7 0

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}

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