Question

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

Answer 1

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)`