What is the derivative of #x^2+y^2#?

1 Answer
Steve M
Feb 7, 2018

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

# d/dy( x^2+y^2) = 2y + 2x' #

Explanation:

If we differentiate wrt #x#, and apply the chain rule. then we get:

# d/dx( x^2+y^2) = d/dx(x^2)+dy/dxd/dy(y^2) #
# " " = 2x + 2y' #

Similarly, we could differentiate wrt #y#

# d/dy( x^2+y^2) = 2y + 2x' #

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