What is the derivative of $f (t) = (2 t - 3 t e^{t}, 2 t^{2} - t)$ $f(t) = (2t-3te^t, 2t^2-t )$ ?

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What is the derivative of $f (t) = (2 t - 3 t e^{t}, 2 t^{2} - t)$ $f(t) = (2t-3te^t, 2t^2-t )$ ?

1 Answer

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Answer 1 · 2018-03-11T14:35:48

$\frac{d f}{d x} = \frac{4 t - 1}{2 - 3 e^{t} - 3 t e^{t}}$ $(df)/(dx)=(4t-1)/(2-3e^t-3te^t)$

Explanation:

In a parametric equation $f (t) = (x (t), y (t))$ $f(t)=(x(t),y(t))$ , the derivative $\frac{d f}{d x} = \frac{\frac{d y}{d t}}{\frac{d y}{d t}}$ $(df)/(dx)=((dy)/(dt))/((dy)/(dt))$

Here we have $f (t) = (2 t - 3 t e^{t}, 2 t^{2} - t)$ $f(t)=(2t-3te^t,2t^2-t)$

As $x (t) = 2 t - 3 t e^{t}$ $x(t)=2t-3te^t$ , $\frac{d x}{d t} = 2 - 3 e^{t} - 3 t e^{t}$ $(dx)/(dt)=2-3e^t-3te^t$

and as $y (t) = 2 t^{2} - t$ $y(t)=2t^2-t$ , $\frac{d y}{d t} = 4 t - 1$ $(dy)/(dt)=4t-1$

and hence $\frac{d f}{d x} = \frac{4 t - 1}{2 - 3 e^{t} - 3 t e^{t}}$ $(df)/(dx)=(4t-1)/(2-3e^t-3te^t)$

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What is the derivative of $f (t) = (2 t - 3 t e^{t}, 2 t^{2} - t)$ $f(t) = (2t-3te^t, 2t^2-t )$ ?

1 Answer

Explanation:

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What is the derivative of f(t)=(2t−3tet,2t2−t)f(t) = (2t-3te^t, 2t^2-t ) ?

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Explanation:

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What is the derivative of $f (t) = (2 t - 3 t e^{t}, 2 t^{2} - t)$ $f(t) = (2t-3te^t, 2t^2-t )$ ?