How do you differentiate the following parametric equation: # x(t)=-te^t-2t, y(t)= 3t^3-4t #? Calculus Parametric Functions Derivative of Parametric Functions 1 Answer Narad T. Jan 14, 2017 The answer is #=-((e^t+te^t+2))/((9t^2-4))# Explanation: We need #(uv)'=u'v-uv'# #(e^x)'=e^x# #(x^n)'=nx^(n-1)# #x(t)=-te^t-2t# #y(t)=3t^3-4t# #dx/dt=-(e^t+te^t+2)# #dy/dt=9t^2-4# #dy/dx=(dy/dt)/(dx/dt)=-((e^t+te^t+2))/((9t^2-4))# 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 1536 views around the world You can reuse this answer Creative Commons License