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How do you find #(dy)/(dx)# given #2x^2+y^2=4#?

Answer 1

Emily Ammerman

#d/dx(2x^2 + y^2) = d/dx(4)#

#4x + 2y(dy/dx) = 0#

#2y(dy/dx) = -4x#

#dy/dx = (-4x)/(2y)#

#dy/dx = (-2x)/y#

Hopefully this helps!

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Answer 2

William Abdalla

To find ( \frac{dy}{dx} ) given the equation ( 2x^2 + y^2 = 4 ), you can implicitly differentiate both sides of the equation with respect to ( x ). After differentiating, solve for ( \frac{dy}{dx} ).

Starting with the given equation:

[ 2x^2 + y^2 = 4 ]

Differentiate both sides with respect to ( x ):

[ \frac{d}{dx}(2x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(4) ]

[ 4x + 2y\frac{dy}{dx} = 0 ]

Now solve for ( \frac{dy}{dx} ):

[ 2y\frac{dy}{dx} = -4x ]

[ \frac{dy}{dx} = \frac{-4x}{2y} ]

[ \frac{dy}{dx} = \frac{-2x}{y} ]

So, ( \frac{dy}{dx} = \frac{-2x}{y} ).

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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.