How do you differentiate the following parametric equation: # x(t)=1-lnt, y(t)= cost-t^2 #? Calculus Parametric Functions Derivative of Parametric Functions 1 Answer bp Feb 4, 2016 #dx/dt = -1/t# and #dy/dt = -sin t -2t# Explanation: #dx/dt = -1/t# and #dy/dt = -sin t -2t#. If required to have #dy/dx# , it would be #dy/dt/dx/dt# = t(sin t +2t) 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 1644 views around the world You can reuse this answer Creative Commons License