How do you differentiate the following parametric equation: # x(t)=t-cos2t, y(t)=tsin^2t #? Calculus Parametric Functions Derivative of Parametric Functions 1 Answer 冠廷 李. Jun 20, 2016 #x'(t)=1+sin2t*2##=1+2sin2t# #y'(t)=sin^2t+t*2costsint# Explanation: #d/dx-cos2t=-(-sin2t)*2=2sin2t# #x'(t)=1+sin2t*2##=1+2sin2t# #d/dxsin^2t=d/dx(sint*sint)=2costsint# #y'(t)=sin^2t+t*2costsint# 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 1499 views around the world You can reuse this answer Creative Commons License