x^2 = ln(xy)
x^2 = lnx + lny
2x = 1/x + 1/y y'
y' = y(2x - 1/x)
y'(1) = e(2-1) = e
or, more explicitly,
ln(xy) = x^2
xy = e^(x^2)
y = e^(x^2)/x
y' = e^(x^2) (2x-1)/x^2
y'(1) = e^1 (2-1)/1 = e
Help!!!
determine the slope of the graph x^2=ln(xy) at point (1,e)
2 answers
Thank you!!!!!!!