How do you differentiate the following parametric equation: # x(t)=(lnt)^2, y(t)= -2t^2-2t^3e^(t) #? Calculus Parametric Functions Derivative of Parametric Functions 1 Answer Shiva Prakash M V Feb 18, 2018 #dy/dx=(-2t^2)/lnt(4+2t^2e^t(t+3))# Explanation: #dy/dx=(dy/dt)/(dx/dt)# #x=(lnt)^2# #dx/dt=2(lnt)(1/t)=2/tlnt# #y=-2t^2-2t^3e^t# #d/dt(e^tf(t)=e^t(f(t)+f'(t))# #f(t)=t^3, f'(t)=3t^2# #dy/dt=-2xx2t-2xx(e^t(t^3+3t^2))=-4t-2e^t(t^3+3t^2)# #dy/dx=(-4t-2e^t(t^3+3t^2))/(2/tlnt)# #dy/dx=-(t(4+2t^2e^t(t+3))/(2/tlnt))# #dy/dx=(-2t^2)/lnt(4+2t^2e^t(t+3))# 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 2386 views around the world You can reuse this answer Creative Commons License