#y= 2xsqrt(9-x^2)#
First let's find the domain of #y#
#y in RR# where #(9-x^2)>= 0 -> absx<= 3#
Hence the domain of #y# is: #-3<=x<=+3#
Next we will find the turning points of #y# using #y'=0#
#y = 2x(9-x^2)^(1/2)#
Applying the product rule and chain rule
#y'# = #2x*1/2*(9-x^2)^(-1/2) * (-2x) + 2*(9-x^2)^(1/2)#
#= 2(9-x^2)^(1/2) - (2x^2)/((9-x^2)^(1/2))#
#= (2(9-x^2)-2x^2)/(sqrt(9-x^2))#
#= (18-4x^2)/sqrt(9-x^2)#
For critical #y#:
# (18-4x^2)/sqrt(9-x^2)=0 -> 18-4x^2=0#
#2x^2 =9 -> x= +- sqrt(9/2)#
#x= +-3/sqrt2 = +-(3sqrt2)/2#
This question asks for the intervals of #x# of increasing and decreasing #y# using the #y'#. To avoid using the second derivative, it is now helpful to observe the graph of #y# below:
graph{2xsqrt(9-x^2) [-20.28, 20.26, -10.13, 10.15]}
Since we know the turning points are where #x= +-(3sqrt2)/2# and that the domain of #y# is #x: [-3, +3]#
We can see that #y# is increasing for #x: (-(3sqrt2)/2, +(3sqrt2)/2)#
and #y# is decresing for #x: [-3, -(3sqrt2)/2) uu (+(3sqrt2)/2, +3]#
