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