How do you find the second derivative of #sin(2x)#? Calculus Graphing with the Second Derivative Notation for the Second Derivative 1 Answer Wataru Apr 13, 2017 #f''(x)=-4sin 2x# Explanation: Let #f(x)=sin 2x#. By #(sin x)'=cos x# and Chain Rule, #f'(x)=cos 2x cdot(2x)'=2cos 2x# By #(cos x)'=-sin x# and Chain Rule, #f''(x)=-2sin 2x cdot (2x)'=-4sin 2x# I hope that this was clear. Answer link Related questions What is notation for the Second Derivative? What is Leibniz notation for the second derivative? What is the second derivative of #e^(2x)#? How do you find the first, second derivative for #3x^(2/3)-x^2#? What is the second derivative of #y=x*sqrt(16-x^2)#? How do you find the first and second derivative of #(lnx)/x^2#? How do you find the first and second derivative of #lnx^(1/2)#? How do you find the first and second derivative of #x(lnx)^2#? How do you find the first and second derivative of #ln(x^2-4)#? How do you find the first and second derivative of #ln(lnx^2)#? See all questions in Notation for the Second Derivative Impact of this question 12737 views around the world You can reuse this answer Creative Commons License