What is the derivative of #f(t) = (1-e^t , sec(1-t) ) #? Calculus Parametric Functions Derivative of Parametric Functions 1 Answer mason m Dec 30, 2015 #f'(t)=(sec(1-t)tan(1-t))/(e^t)# Explanation: #x'(t)=-e^t# #y'(t)=-sec(1-t)tan(1-t)# The derivative of the parametric function is #dy/dx=(y'(t))/(x'(t))=(-sec(1-t)tan(1-t))/(-e^t)=(sec(1-t)tan(1-t))/(e^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 1641 views around the world You can reuse this answer Creative Commons License