You have to find the derivative:
#f'(x)=(x)'sin^3(x/3)+x*(sin^3(x/3))'#
In this case, the derivative of the trigonometric function is actually a combination of 3 elementary functions. These are:
#sinx#
#x^n#
#c*x#
The way this will be solved is as follows:
#(sin^3(x/3))'=3sin^2(x/3)*(sin(x/3))'=#
#=3sin^2(x/3)*cos(x/3)(x/3)'=#
#=3sin^2(x/3)*cos(x/3)*1/3=#
#=sin^2(x/3)*cos(x/3)#
Therefore:
#f'(x)=1*sin^3(x/3)+x*sin^2(x/3)*cos(x/3)#
#f'(x)=sin^3(x/3)+x*sin^2(x/3)*cos(x/3)#
#f'(x)=sin^2(x/3)*(sin(x/3)+xcos(x/3))#
The derivation of the tangent equation:
#f'(x_0)=(y-f(x_0))/(x-x_0)#
#f'(x_0)*(x-x_0)=y-f(x_0)#
#y=f'(x_0)*x-f'(x_0)*x_0+f(x_0)#
Substituting the following values:
#x_0=π#
#f(x_0)=f(π)=π*sin^3(π/3)=2.0405#
#f'(x_0)=f'(π)=sin^2(π/3)*(sin(π/3)+πcos(π/3))=1.8276#
Therefore, the equation becomes:
#y=1.8276x-1.8276*π+2.0405#
#y=1.8276x-3.7#
In the graph below you can see that at #x=π=3.14# the tangent is indeed increasing and will intersect the y'y axis at #y<0#
graph{x(sin(x/3))^3 [-1.53, 9.57, -0.373, 5.176]}
