math
posted by matthias .
find an equation of the tangent line to the graph of y=e^(2x3) at point (3/2,1)
thanks!

y'(x) = 2e^(2x3)
y'(3/2) = 2
so, you want the line through (3/2,1) with slope 2:
y1 = 2(x3/2)
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