How do you differentiate the following parametric equation: # x(t)=t, y(t)=t #? Calculus Parametric Functions Derivative of Parametric Functions 1 Answer Eddie Jun 23, 2016 1 Explanation: seems little point as #x = t = y# meaning #y = x# so #y' = 1# but #dy/dx = (dy/dt)/(dx/dt) = 1/1# 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 1319 views around the world You can reuse this answer Creative Commons License