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# If $g(x) = x/e^x,$ find $g^{(n)}(x).$

## $g^{(n)}=\frac{\left(-1^{n}\right)(x-n)}{e^{x}}$

Derivatives

Differentiation

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### Video Transcript

it's clear. So when you read here so we're given G of X is equal to X over each. The X prefers gonna find the first derivative, and we're gonna make a pattern out of it. You get one times each the X minus B to the x times X overeat the X square, and we get this is equal to one minus X over each X for a second derivative. After we simplify and everything, you get negative two plus X over each the X and for their derivative, you get three minus ranks over E to the X. So here we see that the denominators are all equal to eat the x No for the numerator. They switch off from positive to negative to positive and the numerator constant is equal to end. So it becomes negative one to the end. Power times X minus and for the numerator. So we got G. The end becomes equal to negative one on times X minus and over each the x

Derivatives

Differentiation

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