How do you find the derivative of #(sin2x)#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer YouDid Feb 26, 2015 We should use the chain rule: #f(g(x)) = f'(g(x))*g'(x)#. In our situation #f(x) = sin(x)# and #g(x) = 2x#. Then #f'(x) = cos(x)# and #g'(x) = 2#. So #f(g(x)) = sin(2x) = f'(g(x))*g'(x) = cos(2x)*2 = 2cos(2x)# Answer link Related questions What is Derivatives of #y=sec(x)# ? What is the Derivative of #y=sec(x^2)#? What is the Derivative of #y=x sec(kx)#? What is the Derivative of #y=sec ^ 2(x)#? What is the derivative of #y=4 sec ^2(x)#? What is the derivative of #y=ln(sec(x)+tan(x))#? What is the derivative of #y=sec^2(x)#? What is the derivative of #y=sec^2(x) + tan^2(x)#? What is the derivative of #y=sec^3(x)#? What is the derivative of #y=sec(x) tan(x)#? See all questions in Derivatives of y=sec(x), y=cot(x), y= csc(x) Impact of this question 7215 views around the world You can reuse this answer Creative Commons License