What is the derivative of $f (t) = (t^{3} - e^{3 t - 1}, t - e^{t})$ $f(t) = (t^3-e^(3t-1) , t-e^t )$ ?

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Answer 1 · 2016-06-29T07:31:29

$\frac{1 - e^{t}}{3 t^{2} - 3 e^{3 t - 1}}$ $(1-e^t)/(3t^2-3e^(3t-1))$

This is a parametric function where $x = t^{3} - e^{3 t - 1}$ $x=t^3-e^(3t-1)$ and $y = t - e^{t}$ $y=t-e^t$

and $\frac{d y}{d t} = 1 - e^{t}$ $(dy)/(dt)=1-e^t$ and $\frac{d x}{d t} = 3 t^{2} - 3 e^{3 t - 1}$ $(dx)/(dt)=3t^2-3e^(3t-1)$

Hence $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $(dy)/(dx)=((dy)/(dt))/((dx)/(dt))$

= $\frac{1 - e^{t}}{3 t^{2} - 3 e^{3 t - 1}}$ $(1-e^t)/(3t^2-3e^(3t-1))$

What is the derivative of f(t)=(t3−e3t−1,t−et)f(t) = (t^3-e^(3t-1) , t-e^t ) ?