How do you find the derivative of #f(x)=8sinxcosx#? Calculus Differentiating Trigonometric Functions Derivatives of y=sec(x), y=cot(x), y= csc(x) 1 Answer sjc Feb 16, 2017 #f'(x)=8(cos^2x-sin^2x)=8cos2x# Explanation: This will need the product rule: #f(x)=uv=>f'(x)=vu'+uv'# #f(x)=8sinxcosx# #u=8sinx=>u'=8cosx# #v=cosx=>v'=-sinx# #:.f'(x)=8cosxxxcosx+8sinx xx(-sinx)# #f'(x)=8(cos^2x-sin^2x)=8cos2x# 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 5610 views around the world You can reuse this answer Creative Commons License