calculus
posted by LI on .
Determine the equation of the tangent line at the indicated coordinate f(x)=e^(2x)*in(8x) for x=2
The equation of the tangent line in slopeintercept form is y= ?

I read that as
f(x) = e^(2x) ln(8x)
f'(x) = 2(e^(2x) ln(8x) + (1/x) e^(2x) , using the product rule
f(2) = .0508
f'(2) = .0924
so we have a slope of .0924 and the point (2,.0508
y = .0924x + b
sub in the point
.0508 = .0924(2) + b
b = .2356
y = .0924x + .2356
(Was expecting "nicer" numbers)