Determine the slope of the graph of x^2 = ln(xy) at the point (1, e).

Given: x^2 = ln(xy)

The solution for this problem is:

e^x^2 = xy 

y = (e^x^2)/x;

f(x) = (e^x^2);

f'(x) = 2x*(e^x^2);

g(x) = x; g'(x) = 1 

[2(x^2)(e^x^2) - (e^x^2)]/x^2 = (e^x^2)(2*(x^2) - 1)/x^2;

plug in x = 1 then this will give us the slope, correct answer which is e.

i really do want to but i just don't know it i'm so sorry : (

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