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Implicit differentiation to find dy/dx at a point on an ellipse

Consider the ellipse defined implicitly by x216+y29=1\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1.

Using implicit differentiation, find dydx\dfrac{dy}{dx} for the curve x216+y29=1\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1, and then evaluate it at the point (0,3)(0,3).

Hint

Differentiate both sides with respect to xx, remembering that yy is a function of xx, so ddx[y2]=2ydydx\dfrac{d}{dx}[y^2] = 2y\dfrac{dy}{dx}. Then solve for dydx\dfrac{dy}{dx} and substitute x=0x=0, y=3y=3.

Step-by-step solution

Differentiate both sides of x216+y29=1\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1 with respect to xx. The left side gives 2x16+2y9dydx\dfrac{2x}{16} + \dfrac{2y}{9}\dfrac{dy}{dx}, using the chain rule on y2y^2, and the right side gives 00 since it's a constant.

This simplifies to x8+2y9dydx=0\dfrac{x}{8} + \dfrac{2y}{9}\dfrac{dy}{dx} = 0. Solving for dydx\dfrac{dy}{dx} gives dydx=−9x16y\dfrac{dy}{dx} = -\dfrac{9x}{16y}.

Substituting x=0x=0 and y=3y=3 into dydx=−9x16y\dfrac{dy}{dx} = -\dfrac{9x}{16y} gives dydx=−9(0)16(3)=0\dfrac{dy}{dx} = -\dfrac{9(0)}{16(3)} = 0.

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