If f(x) = 11x/(2 + x 2), find f '(3) and use it to find an equation of the tangent line to the curve y = 11x/(2 + x 2) at the point (3, 3).

[tex]y = \frac{11x}{(2+ x^{2} )}
y ' = ((2+ x^{2} )11 - 11x(2x))/(2+ x^{2})^2 [/tex]
[tex]y ' = (22- x^{2} )/(2+ x^{2} )^2[/tex]
y '(3) = (22 - 11(9))/(2 + 9)^2 = -77/121 = -7/11 (this is the slope at x = 3)

y - 3 = -7/11(x - 3)     (this is the answer)