We want differentiate #f(x)=x/sin(x)#, therefore we seek
#f'(x)=lim_(h->0)((x+h)/sin(x+h)-x/sin(x))/h#
For later use remember the two quite fundamental limits
#color(orange)(lim_(x->0)sin(x)/x=1)# and #color(green)(lim_(x->0)(1-cos(x))/x=0)#
and the trigonometric identity
#color(brown)(sin(x+h)=cos(x)sin(h)+cos(h)sin(x))#
Let's rewrite our problem
#f'(x)=lim_(h->0)((x+h)/(sin(x+h))-x/sin(x))/h#
#=lim_(h->0)((x+h)sin(x)-xsin(x+h))/(hsin(x+h)sin(x))#
#=lim_(h->0)(hsin(x))/(hsin(x+h)sin(x))+lim_(h->0)(xsin(x)-xsin(x+h))/(hsin(x+h)sin(x))#
#=L_1+L_2#
Let's rewrite the second limit
#L_2=lim_(h->0)(xsin(x)-xcolor(brown)((cos(x)sin(h)+cos(h)sin(x))))/(hsin(x+h)sin(x))#
#=lim_(h->0)(xsin(x)(1-cos(h))-x(cos(x)sin(h)))/(hsin(x+h)sin(x))#
#=lim_(h->0)(xsin(x)(1-cos(h)))/(hsin(x+h)sin(x))-lim_(h->0)(xcos(x)sin(h))/(hsin(x+h)sin(x))#
#=L_3-L_4#
Such that #f'(x)=L_1+L_3-L_4#
What is left to evaluate these limits
Limit 1
#L_1=lim_(h->0)(hsin(x))/(hsin(x+h)sin(x))=csc(x)#
Limit 3
#L_3=lim_(h->0)(xsin(x)(1-cos(h)))/(hsin(x+h)sin(x))#
#=xcolor(green)(lim_(h->0)(1-cos(h))/(h))lim_(h->0)1/sin(x+h)=0#
Limit 4
#L_4=lim_(h->0)(xcos(x)sin(h))/(hsin(x+h)sin(x))#
#=xcos(x)/sin(x)color(orange)(lim_(h->0)(sin(h))/(h))lim_(h->0)1/sin(x+h)=xcot(x)csc(x)#
Remember #f'(x)=L_1+L_3-L_4# so
#f'(x)=csc(x)-xcot(x)csc(x)#
