posted by LI .
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 slope-intercept 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)