How do you differentiate # g(x) =sin^2(x/6) #? Calculus Differentiating Trigonometric Functions Special Limits Involving sin(x), x, and tan(x) 1 Answer Shwetank Mauria Mar 31, 2016 #(dg)/(dx)=1/6sin(x/3)# Explanation: To differentiate #g(x)=sin^2(x/6)#, we use the concept of function of function, as #g(x)=(f(x))^2# where #f(x)=sin(x/6)#. Hence, #(dg)/(dx)=(dg)/(df)*(df)/(dx)# = #2sin(x/6)*cos(x/6)*1/6# = #1/6sin(x/3)# (using formula #sin2x=2sinxcosx#) Answer link Related questions What are Special Limits Involving #y=sin(x)#? How do you find the limit #lim_(x->0)sin(x)/x# ? How do you find the limit #lim_(x->0)tan(x)/x# ? What is the derivative of #tanx^3#? What is the derivative of #tanx/x#? How do you differentiate # g(x) =(1+cosx)/(1-cosx) #? What is the derivative of #tan(2x)#? How do you differentiate #f(x)=sinx/x#? How do you differentiate #f(x)=sinx/(1-cosx)#? How do you differentiate #f(x)=(x+2)/cosx#? See all questions in Special Limits Involving sin(x), x, and tan(x) Impact of this question 2442 views around the world You can reuse this answer Creative Commons License