Let #f(x) = x^2 sin(x)#. Then
#(df)/dx = lim_{h \to 0} (f(x+h) - f(x))/h#
#= lim_{h \to 0} ((x + h)^2sin(x + h) - x^2sin(x))/h#
#= lim_{h \to 0} ((x^2 + 2hx + h^2)(sin(x)cos(h) + sin(h)cos(x)) - x^2sin(x))/h#
#=#
# lim_{h \to 0} (x^2sin(x)cos(h) - x^2sin(x))/h +#
# lim_{h \to 0} (x^2sin(h)cos(x))/h + #
# lim_{h \to 0} (2hx(sin(x)cos(h) + sin(h)cos(x)))/h + #
# lim_{h \to 0} (h^2(sin(x)cos(h) + sin(h)cos(x)))/h#
by a trigonometric identity and some simplifications. On these four last lines we have four terms .
The first term equals 0, since
#lim_{h \to 0} (x^2sin(x)cos(h) - x^2sin(x))/h#
#= x^2sin(x) ( lim_{h \to 0}(cos(h) - 1)/h) #
#= 0#,
which can be seen e.g. from Taylor expansion or L'Hospital's rule.
The Fourth term also vanishes because
#lim_{h \to 0} (h^2(sin(x)cos(h) + sin(h)cos(x)))/h#
#= lim_{h \to 0} h(sin(x)cos(h) + sin(h)cos(x))#
#= 0#.
Now the second term simplifies to
# lim_{h \to 0} (x^2sin(h)cos(x))/h#
#=x^2cos(x)(lim_{h \to 0} (sin(h))/h)#
#= x^2cos(x)#,
since
#lim_{h \to 0} (sin(h))/h = 1#, as shown here, or e.g. L'Hospital's rule (see below).
The third term simplifies to
# lim_{h \to 0} (2hx(sin(x)cos(h) + sin(h)cos(x)))/h#
# = lim_{h \to 0} 2xsin(x)cos(h) + 2xsin(h)cos(x)#
#= 2xsin(x)#,
which after adding to the second term gives that
#(df)/dx = 2xsin(x) + x^2cos(x)#.
Note: By L'Hospital's rule, since #\lim_{h \to 0} sin(h)=0# and #\lim_{h \to 0} h=0# and both functions are differentiable around #h=0#, we have that
#\lim_{h \to 0} sin(h)/h = \lim_{h \to 0} ((d/(dh))sin(h))/(d/(dh) h) = \lim_{h \to 0} cos(h)=1#.
The limit # lim_{h \to 0}(cos(h) - 1)/h =0# can be shown similarly.
