How do you differentiate #f(x) =sin(3x-2)* cos(3x -2) #? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Konstantinos Michailidis Apr 4, 2016 Remember that #sin2A=2*sinA*cosA# Hence #f(x)=sin(3x-2)*cos(3x-2)=>f(x)=1/2*sin(2*(3x-2))# Hence #(df)/dx=1/2*cos(2(3x-2))*6=3*cos(6x-4)=# Finally #f'(x)=3*cos(6x-4)# Answer link Related questions What is the derivative of #f(x)=sin^-1(x)# ? What is the derivative of #f(x)=cos^-1(x)# ? What is the derivative of #f(x)=tan^-1(x)# ? What is the derivative of #f(x)=sec^-1(x)# ? What is the derivative of #f(x)=csc^-1(x)# ? What is the derivative of #f(x)=cot^-1(x)# ? What is the derivative of #f(x)=(cos^-1(x))/x# ? What is the derivative of #f(x)=tan^-1(e^x)# ? What is the derivative of #f(x)=cos^-1(x^3)# ? What is the derivative of #f(x)=ln(sin^-1(x))# ? See all questions in Differentiating Inverse Trigonometric Functions Impact of this question 2402 views around the world You can reuse this answer Creative Commons License