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I am curious, is the problem 6e^(2y-3x) with the minus 3x in the exponent? or is it (6e^2y)-3x?
because really all that you have to do is take the derivative of both side to get your slope then after you have that you plug the point and the slope into the equation for a line and that will give you your tangent line
xy=6e^(2y-3x) at P(3,2).
I'm just having trouble with the implicit differentiation of e's exponent. That's troubling me a little bit.
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the derivative of e^u, using the chain rule is (e^u)u'. so when you take the derivative yo get (e^2y-3x)(2dy/dx-3) Now you want to solve for dy/dx and that is your derivative. so your equation will look like this
soved for dy/dx
you might want to double check my algebra on that
now plg in you values for x and y at that point and that will give you the slope at that point