What is the derivative of #f(t) = (t^2+1 , t-e^(1-t) ) #? Calculus Parametric Functions Derivative of Parametric Functions 1 Answer turksvids Jan 3, 2018 #f'(t)= (2t, 1+e^(1-t))# Explanation: Take the derivative component-by-component: #f(t) = (t^2+1, t-e^(1-t))# #f'(t) = (d/dt(t^2+1), d/dt(t-e^(1-t)))# #= (2t, 1-e^(1-t)(-1))# (don't forget the chain rule!) #f'(t)= (2t, 1+e^(1-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 1179 views around the world You can reuse this answer Creative Commons License