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

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)

CO
Answered by Callum O. Maths tutor

43349 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Two particles, A and B, are moving directly towards each other on a straight line with speeds of 6 m/s and 8 m/s respectively. The mass of A is 3 kg, and the mass of B is 2 kg. They collide to form a single particle of speed "v" m/s. Find v.


Find the derivation of (sinx)(e^2x)


In this question, take 'log' to mean 'log base 5'. Solve the equation log(x^2-5)-log(x) = 2*log(2)


y = 2/x^3 find and expression for dy/dx


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning