What is the derivative of #f(t) = (t^3-e^(1-t) , tan^2t ) #? Calculus Parametric Functions Derivative of Parametric Functions 1 Answer Shwetank Mauria Mar 31, 2016 #(dy)/(dx)=(3e^2+e^(1-t))/(2tant*sec^2t)#. Explanation: #f(t)=(t^3-e^(1-t),tan^2t)# is a parametric function in which both #x# and #y# are function of #t#. As such #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#. Here #(dy)/(dt)=3e^2-(-1)*e^(1-t)=3e^2+e^(1-t)# and #(dx)/(dt)=2tant*sec^2t# Hence #(dy)/(dx)=(3e^2+e^(1-t))/(2tant*sec^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 1404 views around the world You can reuse this answer Creative Commons License