How do you find the derivative of $e^(x-1)$ ?

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Answer 1 · 2016-09-05T12:37:30

$e^(x - 1)$

We have: $e^(x - 1)$

This expression can be differentiated using the "chain rule".

Let $u = x - 1 => u' = 1$ and $v = e^(u) => v' = e^(u)$ :

$=> (d) / (dx) (e^(x - 1)) = 1 cdot e^(u)$

$=> (d) / (dx) (e^(x - 1)) = e^(u)$

We can now replace $u$ with $x - 1$ :

$=> (d) / (dx) (e^(x - 1)) = e^(x - 1)$

How do you find the derivative of e^(x-1)e^(x-1)?