# Differentiate with respect to X: x^2 + 2y^2+ 2xy = 2

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Assuming the correct tools of differentiation have been taught, we can tackle each term seperately and then rearrange to have dy/dx as the subject.

Taking a look at the first term, x^2,  differentiating this term would become 2x (diffentiating x^n = nx^n-1)

Taking a look at the second term, 2y^2, it would appear we could differentiate it just like we did the first term. However this variable involves y and not x, meaning we must differentiate it implicitly.Therefore differentiating 2y^2 would become 4y(dy/dx)

Taking a look at the third term, 2xy, we immediately notice that it has both x terms and y terms involved; this should immediately hint to us that the product rule should be used. Therefore differentiating 2xy would become 2y + 2x(dy/dx) (Differentiating any term involving any other variable other than x with respect to x would require implicit differentiation).

Differentiating any constant (2) would = 0

Putting all these terms together would give:

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

With our basic GCSE knowledge of subject formula we can get:

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

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

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