Suppose y = ln(x^2+y^2). Find: y'(1)

Answer 1

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

y'= (1/(x^2+y^2)(2x+2yy')= (2/(x^2+y^2)(x+yy')

(x^2+y^2)/2 = (x+yy')/y' = x/y' + y

x/y' = (x^2+y^2)/2 - y= (x^2+y^2 -2y)/2

y'/x = 2/(x^2+y^2-2y)

y'(x,y) = 2x/(x^2 + y^2 -2y)

y'(1,y) = 2/(1+y^2 -2y)

if you treat x^2+y^2 = 1, as a circle, then

x=1 means y=0, and

y'(x,y) = y'(1,0) = 2/(1) = 2

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

e^y = x^2 + y^2

e^0= 1 = 1^2 + (0)^2 = 1

0=ln(1^2+ 0^2)= ln1 = 0

BUT IF, in the unlikely event

you meant find y'(1) = y'(x,1) with y=1 then y=ln(x^2+y^2) becomes 1=ln(x^2+1)

and e^1 = e = x^2+1

x^2 =e-1

x = + or - sqr(e-1)

and y'(sqr(e-1), 1) = 2/(x+y^2-2y)= 2/(sqr(e-1) +1-2)

=2/(sqr(e-1)-1) = about 3.0831= about 3, to nearest integer

or y'(-sqr(e-1), 1) = 2/(-sqr(e-1) -1) = about -0.76

Answer 2

y' = (1/(x²+y²)(2x+2yy') By chain rule

Isolate y'

y'(1-2y/(x²+y²)= 2x/(x²+y²)

y'(x²+y²-2y) = 2x

y'= 2x/(x²+y²-2y)