Derivative of Parametric Functions
Key Questions
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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.
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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 andx separately, then place the result ofdy/dt overdx/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)) -
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)} .