Question #5cb24 Calculus Parametric Functions Derivative of Parametric Functions 1 Answer Eddie Jul 6, 2016 #1/2# Explanation: #dy/dx = (dy/dt)/(dx/dt) = (2e^{2t})/(2e^t) = e^t# #(d^2y)/dx ^2= d/dx e^t= d/dt e^t dt/dx =e^t 1/(2 e^t) = 1/2# 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 1167 views around the world You can reuse this answer Creative Commons License