# calculus

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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= ?

• calculus -

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)

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