How do you differentiate the following parametric equation: # x(t)=(t+1)e^t, y(t)= (t-2)^2+t#? Calculus Parametric Functions Derivative of Parametric Functions 1 Answer Aaditya Raghavan Jul 26, 2017 #(2t-3)/(e^t t + 2e^t)# Explanation: #dy/dx = dy/dt / dx/dt# since #dy/dt / dx/dt = dy/dt * dt/dx = dy/dx# (cancelling #dt# in the numerator and denominator) #x(t) = (t+1)e^t# #dx/dt = d/dt((t+1)e^t) = (t+1)e^t + e^t# #y(t) = (t-2)^2 + t# #dy/dt = d/dt((t-2)^2 + t) = 2(t-2) + 1# Therefore, #dy/dx = (2(t-2) + 1)/((t+1)e^t + e^t) = (2t-3)/(e^t t + 2e^t)# 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 1419 views around the world You can reuse this answer Creative Commons License