How do you find the derivative of #1/2sin2x#? Calculus Basic Differentiation Rules Product Rule 1 Answer sjc Apr 5, 2018 #(dy)/(dx)=cos2x# Explanation: we will use the chain rule #(dy)/(dx)=color(red)((dy)/(du))(du)/(dx)# #y=1/2sin2x# #u=2x=>(du)/(dx)=2# #color(red)(y=1/2sinu=>(dy)/(du)=1/2cosu)# #(dy)/(dx)=color(red)(1/2cosu ) xxx2# #(dy)/(dx)=cosu=cos2x# Answer link Related questions What is the Product Rule for derivatives? How do you apply the product rule repeatedly to find the derivative of #f(x) = (x - 3)(2 - 3x)(5 - x)# ? How do you use the product rule to find the derivative of #y=x^2*sin(x)# ? How do you use the product rule to differentiate #y=cos(x)*sin(x)# ? How do you apply the product rule repeatedly to find the derivative of #f(x) = (x^4 +x)*e^x*tan(x)# ? How do you use the product rule to find the derivative of #y=(x^3+2x)*e^x# ? How do you use the product rule to find the derivative of #y=sqrt(x)*cos(x)# ? How do you use the product rule to find the derivative of #y=(1/x^2-3/x^4)*(x+5x^3)# ? How do you use the product rule to find the derivative of #y=sqrt(x)*e^x# ? How do you use the product rule to find the derivative of #y=x*ln(x)# ? See all questions in Product Rule Impact of this question 11169 views around the world You can reuse this answer Creative Commons License