What is the derivative of #y=sqrt(x^2+3y^2)#?

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Answer 1 · 2016-10-29T22:02:50

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

Although you could differentiate it in it's present form using # y=(x^2+3y^2)^(1/2) #, it is much easier to manipulate it as follows:

# y = sqrt(x^2+3y^2) => y^2 = x^2+3y^2 #
# :. x^2+2y^2 = 0 #

We can now very easily differentiate this implicitly and we have the advantage of not having messy #1/2# powers to deal with:

Differentiating wrt #x# gives us:

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

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

# :. x + dy/dxd/dy(y^2) = 0 # (this is the implicit differentiation)

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

# :. x+2ydy/dx = 0 #
# :. dy/dx = -x/(2y) #