How do you differentiate #y=sin x^2#? Calculus Differentiating Trigonometric Functions Differentiating sin(x) from First Principles 1 Answer Alan N. Mar 7, 2017 #dy/dx=2xcos(x^2)# Explanation: #y = sin(x^2)# Applying the chain rule: #dy/dx= cos(x^2) * d/dx(x^2)# #= cos(x^2) * 2x# [Power rule] #= 2xcos(x^2)# Answer link Related questions How do you differentiate #f(x)=sin(x)# from first principles? What is the derivative of #y=3sin(x) - sin(3x)#? How do you find dy/dx if #x + tan(xy) = 0#? How do you find the derivative of the function #y=cos((1-e^(2x))/(1+e^(2x)))#? How do you differentiate #f(x)=2secx+(2e^x)(tanx)#? How do you find the derivate for #y = pisinx - 4cosx#? How do you find the derivative of #f(t) = t^2sin t#? What is the derivative of #sin^2(lnx)#? How do you compute the 200th derivative of #f(x)=sin(2x)#? How do you find the derivative of #sin(x^2+1)#? See all questions in Differentiating sin(x) from First Principles Impact of this question 210676 views around the world You can reuse this answer Creative Commons License