Differentiating #f(x)# is determined by applying the product
#" "#
differentiation on the functions #2x^5+3" "and cosx^2#,and
#" "#
applying chain rule on #cosx^2#.
#" "#
Product differentiation:
#" "#
#color(blue)((u(x)xxv(x))'=u'(x)xxv(x)+v'(x)xxu(x)#
#" "#
Let us differentiate #2x^5 +3# using polynomial differentiation.
#" "#
#color(purple)((2x^5+3)')=(2x^5)'+(3)' =10x^4+0=color(purple)(10x^4)#
#" "#
#" "#
Let us differentiate #cosx^2# by applying chain rule.
#" "#
Chain Rule:
#" "#
#color(red)((u(v(x))'=u'(v(x))xxv'(x))#
#" "#
#color(red)((cos(x^2))'=cos'(x^2)xx(x^2)'=-sinx^2xx(2x)#
#" "#
#color(red)((cos(x^2))'-2xsinx^2)#
#" "#
Differentiating #f(x)# by applying the product differentiation above:
#" "#
#f'(x)=color(blue)((2x^5+3)'xxcosx^2+(cosx^2)'xx(2x^5+3))#
#" "#
#f'(x)=color(purple)(10x^4)xxcosx^2+(color(red)(-2xsinx^2))xx(2x^5+3)#
#" "#
#f'(x)=10x^4cosx^2-4x^6sinx^2-6xsinx^2#
