What is the derivative of #f(t) = (tsint , t^2-t tant ) #?

Question

What is the derivative of #f(t) = (tsint , t^2-t tant  )   #?

Samantha Adkison · Accepted Answer

#(dy)/(dx)=(2t-tant-tsec^2t)/(sint+tcost)# For parametric form of equation, #(dy)/(dx)=((dy)/(dt))/((dx)/(dt))#. Here as #x=tsint#, #(dx)/(dt)=1xxsint+tcost=sint+tcost# (using product rule) and as #y=t^2-t tant#, #(dy)/(dt)=2t-tant-tsec^2t# (using product rule) hence #(dy)/(dx)=(2t-tant-tsec^2t)/(sint+tcost)#

Aurora Alvarado · Answer

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$$
\frac{d}{dt} (t \sin t) = \sin t + t \cos t
$$

$$
\frac{d}{dt} (t^2 - t \tan t) = 2t - (\tan t + t \sec^2 t)
$$

So, the derivative of $ f(t) $ is:

$$
f'(t) = \left( \sin t + t \cos t, \, 2t - \tan t - t \sec^2 t \right)
$$