Use the first derivative to determine where the graph of y = x/x^2+1 is rising. Help!?

To find where the function

#y = x/(x^2+1)#

is increasing, we first take the derivative of it:

Use the quotient rule, which states

#d/(dx)[u/v] = (v(du)/(dx) - u(dv)/(dx))/(v^2)#

where

• #u = x#

• #v = x^2+1#:

#y'(x) = ((x^2+1)d/(dx)[x] - xd/(dx)[x^2+1])/((x^2+1)^2)#

#y'(x) = (x^2 + 1 - 2x^2)/((x^2+1)^2)#

#color(blue)(y'(x) = (1-x^2)/((x^2+1)^2)#

Setting the numerator equal to #0# gives us that #y'(x) = 0# when #x = +-1# (these are the local extrema of the original function).

To find where the graph increases, we look at the three intervals we have:

#(-oo, -1)# (before first local extreme)

#(-1, 1)# (in between local extremes)

#(1, oo)# (after second local extreme)

Plug in one #x# value from each of these three intervals to see if the interval yields a positive number or a negative number; A positive number indicates the function is increasing during that interval. I'll choose #-2#, #0#, and #2#:

#y'(-2) = (1-(-2)^2)/(((-2)^2+1)^2) = -3/25# #larr " negative"#

#y'(0) = (1-(0)^2)/(((0)^2+1)^2) = 1# #larr color(green)(" positive"#

#y'(2) = (1-(2)^2)/(((2)^2+1)^2) = -3/25# #larr " negative"#

Thus, the function is increasing only on the interval

#color(red)(ul(x in (-1, 1)#