What is the derivative of $f (t) = (e^{t^{2} - 1} - e^{t}, 2 t^{2} - 4 t)$ $f(t) = (e^(t^2-1)-e^t, 2t^2-4t )$ ?

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Answer 1 · 2016-07-08T07:53:36

Knowing the chain rule for derivatives, you can see that

$\frac{d x}{d t} = 2 t e^{t^{2} - 1} - e^{t}$ $dx/dt=2te^(t^2-1)-e^(t)$

and

$\frac{d y}{d t} = 4 t - 4$ $dy/dt=4t-4$

Thus, $f'(t)=(2te^(t^2-1)-e^(t), 4t-4)$

In this case, the "trick" is to apply the chain rule to the the first term of the expression

$e^(t^2-1)-e^t$

to obtain the derivative.

The chain rule formula used in this problem comes from

$d/dx[e^(z)]=e^(z) dz/dx$

where

$z=t^2-1$

What is the derivative of f(t) = (e^(t^2-1)-e^t, 2t^2-4t ) f(t)=(et2−1−et,2t2−4t)f(t) = (e^(t^2-1)-e^t, 2t^2-4t ) ?