Derivative of Parametric Functions
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Key Questions

For the parametric equations
#{(x=x(t)),(y=y(t)):}# ,we can find the derivative
#{dy}/{dx}={{dy}/{dt}}/{{dx}/{dt}}={y'(t)}/{x'(t)}# . 
Let
#{(x=x(t)),(y=y(t)):}# .First Derivative
#{dy}/{dx}={{dy}/{dt}}/{{dx}/{dt}}={y'(t)}/{x'(t)}# Second Derivative
#{d^2y}/{dx^2}=d/{dx}[{y'(t)}/{x'(t)}]=1/{{dx}/{dt}}{d}/{dt}[{y'(t)}/{x'(t)}]# by Quotient Rule,
#=1/{x'(t)}cdot{y''(t)x'(t)y'(t)x''(t)}/{[x'(t)]^2}# #={y''(t)x'(t)y'(t)x''(t)}/{[x'(t)]^3}# I hope that this was helpful.

To find the derivative of a parametric function, you use the formula:
#dy/dx = (dy/dt)/(dx/dt)# , which is a rearranged form of the chain rule.To use this, we must first derive
#y# and#x# separately, then place the result of#dy/dt # over#dx/dt# .
#y=t^2 + 2# #dy/dt = 2t# (Power Rule)
#x=tsin(t)# #dx/dt = sin(t) + tcos(t)# (Product Rule)
Placing these into our formula for the derivative of parametric equations, we have:
#dy/dx = (dy/dt)/(dx/dt) = (2t)/(sin(t)+tcos(t))#