What is the derivative of #sin (x/2)#? Calculus Differentiating Trigonometric Functions Intuitive Approach to the derivative of y=sin(x) 1 Answer Eddie Jun 21, 2016 #1/2 cos (x/2)# Explanation: if you know that #d/dx sin x = cos x# then you can use the chain rule so by letting #u = x / 2# you have #d/(du) sin u = cos u# and #(du)/dx = (d)/dx (x/2)= 1/2# #so d/{du} (sin u) \times (du)/(dx) = cos u \times 1/2 = 1/2 cos (x/2)# Answer link Related questions What is the derivative of #-sin(x)#? What is the derivative of #sin(2x)#? How do I find the derivative of #y=sin(2x) - 2sin(x)#? How do you find the second derivative of #y=2sin3x-5sin6x#? How do you compute #d/dx 3sinh(3/x)#? How do you find the derivative #y=xsinx + cosx#? What is the derivative of #sin(x^2y^2)#? What is #f'(-pi/3)# when you are given #f(x)=sin^7(x)#? How do you find the fist and second derivative of #pi*sin(pix)#? If f(x)= 2x sin(x) cos(x), how do you find f'(x)? See all questions in Intuitive Approach to the derivative of y=sin(x) Impact of this question 2896 views around the world You can reuse this answer Creative Commons License