Differentiating #f (x)# is determined by applying the product rule,
#" "#
Chain rule ,and quotient rule.
#" "#
Product rule:
#" "#
#color (blue)((d(uv))/(dx)=(du)/(dx)xx v+ (dv)/(dx)xx u)#
#" "#
Chain rule says:
#" "#
#color (brown)((d (v (u (x))))/(dx)= (d (u(x)))/(dx) xx (dv)/(du))#
#" "#
Quotient rule :
#" "#
#color (green)(( ((u (x))/(v (x))))/(dx) = ((du (x))/(dx) xx v (x) - (dv (x))/(dx) xx u (x))/(v (x))^2#
#" "#
Let's start differentiating #f (x)# by applying the product rule first.
#" "#
#color (blue)((df (x))/(dx)= (dx)/(dx) xx sin (1/x) + (d (sin (1/x)))/(dx) xx x)#
#" "#
Second we should apply Chain rule on# (d (sin (1/x)))/(dx)#
#" "#
#(d f (x))/(dx)= (dx)/(dx) xx sin (1/x) + (color (brown)((d (1/x))/(dx) xx (d (sin (1/x)))/(d (1/x)))) xx x#
#" "#
Now , applying the quotient rule on#(d (1/x))/(dx)#
#" "#
#(d f (x))/(dx)= (dx)/(dx) xx sin (1/x) + (color (green)((0 xx x -(dx)/(dx)xx 1)/(x^2))xx (d(sin (1/x)))/(d (1/x)))) xx x#
#" "#
#(d f (x))/(dx)= 1 xx sin (1/x) + (-1/x^2) xx cos (1/x) xx x#
#" "#
#(d f (x))/(dx)= sin (1/x) -1/x xx cos (1/x)#
