How do you differentiate $y = e^{1 - x}$ $y= e^(1-x)$ ?

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Answer 1 · 2016-06-26T17:36:18

$y' = - e^(1-x)$

$y=e^(1-x) = e*e^(-x).$
$y' = (e*e^(-x)) '=e (e^(-x)) ' = e(-e^(-x))$
$= -( e*e^(-x))$
$= -( e^(1-x))$

or using logs

$y=e^(1-x)$
$ln y=1-x$
$1/y y' = -1$
$y' = -y = - e^(1-x)$

slopes okay.
45° at y' = 1 looks okay.
$y'-2.718281828*((2.718281828)^((x)^-2))$ .
areas don't look okay

graph{(y-2.718281828/(2.718281828^x))=0 [-1, 4, -0.5, 2]}.

graph{(y-2.718281828*((2.718281828)^x^(-2)))=0 [-1, 4, -0.5, 2]}

How do you differentiate y= e^(1-x) y=e1−xy= e^(1-x) ?