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How do you use implicit differentiation to find #(dy)/(dx)# given #x^2+xy+y^2=9#?

Answer 1

Luke Alvarez

#y' = -2(x+y)/(x+2y)#

Differentiate the equation with respect to #x#, considering that based on the chain rule:

#d/(dx) f(y) = (df)/(dy)*(dy)/(dx)#

Thus we have:

#d/(dx) (x^2+xy+y^2) = 0#

#2x+ xy'+y+2yy'=0#

solving for #y'#:

#y'(x+2y) = -2x-y#

#y' = -2(x+y)/(x+2y)#

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Answer 2

Charles Anctil

To use implicit differentiation to find ( \frac{dy}{dx} ) given ( x^2 + xy + y^2 = 9 ):

Differentiate both sides of the equation with respect to (x).
Treat (y) as a function of (x) and use the chain rule when differentiating terms involving (y).
Solve for ( \frac{dy}{dx} ) after differentiation.

The steps are as follows:

[ x^2 + xy + y^2 = 9 ] [ \frac{d}{dx}(x^2) + \frac{d}{dx}(xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(9) ]

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

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

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

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

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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.