How do you find the derivative of #y=ln (x^2+y^2)#?

1 Answer
Steve M
Sep 15, 2017

We have:

# y = ln(x^2+y^2) #

Method 1: Implicit differentiation, as is:

Using the chain rule:

# dy/dx = 1/(x^2+y^2)(2x+2ydy/dx) #
# " " = (2x)/(x^2+y^2) + (2y)/(x^2+y^2)dy/dx #

# :. (1 - (2y)/(x^2+y^2))dy/dx = (2x)/(x^2+y^2) #

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

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

Method 2: Use exponentials:

# y = ln(x^2+y^2) iff e^y = x^2+y^2 #

Implicit differentiation yields:

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

Method 3: Implicit Function Theorem

Putting:

# F(x,y) = ln(x^2+y^2) -y #

We have:

# (partial F)/(partial x) = (2x)/(x^2+y^2) #
# (partial F)/(partial y) = (2y)/(x^2+y^2) -1 = -(x^2+y^2-2y)/(x^2+y^2)#

Then:

# dy/dx = - ((partial F)/(partial x)) / ((partial F)/(partial y)) #
# \ \ \ \ \ \ = - ((2x)/(x^2+y^2)) / (-(x^2+y^2-2y)/(x^2+y^2)) #
# \ \ \ \ \ \ = (2x) / (x^2+y^2-2y) #

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