What is the slope of the line tangent to the graph of #x^2xy+y^2=7#?

Answer 1

Cameron Alexander

Slope= #(-3x^2 y)/(2y+x^3)#

Differentiating the equation, it is #x^3 dy/dx +3x^2 y +2y dy/dx=0#

#(x^3 +2y)dy/dx= -3x^2y#

#dy/dx= (-3x^2 y)/ (2y+x^3)#

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Answer 2

Emma Aldridge

To find the slope of the line tangent to the graph of (x^2xy + y^2 = 7), we differentiate the equation implicitly with respect to (x), then solve for (\frac{dy}{dx}).

Differentiating both sides with respect to (x) yields: [2x^2y + 2xy + x^2\frac{dy}{dx} + 2y\frac{dy}{dx} = 0]

Next, isolate (\frac{dy}{dx}): [\frac{dy}{dx}(x^2 + 2y) = -2x^2y - 2xy]

Divide both sides by (x^2 + 2y): [\frac{dy}{dx} = \frac{-2x^2y - 2xy}{x^2 + 2y}]

This expression represents the slope of the tangent line at any point on the curve.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.