a)Find dy/x where y=tan(4=x^3)^4?
B) Find relative extrema of f(x)=x^3-3x^2+4
y = u^4 where u = tan(4+x^3)
y' = 4u^3 u'
u = tan(v) where v = 4+x^3
y' = 4u^3 (sec^2(v) v')
= 4 tan^3(4+x^3) * sec^2(4+x^3) * 3x^2
= 12x^2 tan^3(4+x^3) sec^2(4+x^3)
y = x^3 - 3x^2 + 4
y' = 3x^2 - 6x = 3x(x-2)
y'=0 when x=0,2
y(0) = 4
y(2) = 0
so, extrema are at (0,4) and (2,0)