How do you find the derivative of # sin^2(x/6)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Konstantinos Michailidis May 13, 2016 Well it is #d(sin^2(x/6))/dx=2sin(x/6)*d(sin(x/6))/dx= 2*sin(x/6)*cos(x/6)*(x/6)'= 1/3*sin(x/6)*cos(x/6)= (1/(2*3))*sin(2*x/6)= 1/6 * sin(x/3)# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1056 views around the world You can reuse this answer Creative Commons License