How do you differentiate the following parametric equation: # (3sin(t/6-pi/12), 2tcos(pi/3-t/8))#? Calculus Parametric Functions Derivative of Parametric Functions 1 Answer Pankaj Solanki Sep 18, 2017 #dx/dt=1/2 cos(t/6-pi/12)# #dy/dt=2{cos(pi/3-t/8)+t/8sin(pi/3-t/8)}# Explanation: #x=3sin(t/6-pi/12)# #dx/dt=3(1/6)cos(t/6-pi/12)=1/2 cos(t/6-pi/12)# #y=2tcos(pi/3-t/8)# #dy/dt=2{cos(pi/3-t/8)d/dt(t)+td/dt(cos(pi/3-t/8)}# #dy/dt=2{cos(pi/3-t/8)*(1)-t(-1/8)sin(pi/3-t/8)}# #dy/dt=2{cos(pi/3-t/8)+t/8sin(pi/3-t/8)}# Answer link Related questions How do you find the second derivative of a parametric function? How do you find derivatives of parametric functions? How do you find #dy/dx# for the curve #x=t*sin(t)#, #y=t^2+2# ? How do you find the equation of the tangent to the curve #x=t^4+1#, #y=t^3+t# at the point... How do you find #(d^2y)/(dx^2)# for the curve #x=4+t^2#, #y=t^2+t^3# ? How do you find parametric equations of a tangent line? How do you find parametric equations for the tangent line to the curve with the given parametric... How do you find the equation of a line tangent to the curve at point #t=-1# given the parametric... How do you differentiate the following parametric equation: # x(t)=t^3-5t, y(t)=(t-3) #? How do you differentiate the following parametric equation: # x(t)=lnt, y(t)=(t-3) #? See all questions in Derivative of Parametric Functions Impact of this question 1649 views around the world You can reuse this answer Creative Commons License