f'(x) = 1/2 (2 - 1/x^2) e^x^2
So plug in x=1
Looks like 1/2 e^2
Determine the slope of the tangent at the indicated value of x. State as an exact value.
f(x)= e^x^2 / 2x, x=1
3 answers
using the quotient rule:
f '(x) = (2x(2x)e^(x^2) - 2 e^(x^2) )/(4x^2)
= ( e^(x^2) (4x^2 - 2x) )/(4x^2)
when x = 1
f '(1) = e^1 (2)/4
= (1/2)e or e/2
f '(x) = (2x(2x)e^(x^2) - 2 e^(x^2) )/(4x^2)
= ( e^(x^2) (4x^2 - 2x) )/(4x^2)
when x = 1
f '(1) = e^1 (2)/4
= (1/2)e or e/2
duh - saved again.
0 answers