Question

if g(x)=x/e^x find g^(n)(x)

Answer 1

@zepdrix

Answer 2

Take the first derivative, then try to simplify it before we take a second derivative.
We should start to see a pattern by the second derivative.

What do you get for g'(x)? :)

Answer 3

i did g'(x)=e^-x-xe^-x, g"(x)=xe^-x-2e^-x, g^(3)=3e^-x-xe^-x

Answer 4

i tried both e^-x(x-1) and e^-x(1-x). they r both wrong

Answer 5

\[\Large g(x)\quad=\quad \frac{x}{e^x}\]\[\Large g'(x)\quad=\quad \frac{1-x}{e^x}\]Ok your first derivative looks good.
We need to simplify down the second derivative a bit.
You should get something like this:\[\Large g''(x)\quad=\quad \frac{-(2-x)}{e^x}\]

Answer 6

so the answer would be (n-x)/e^x?

Answer 7

So you noticed a pattern right?
Good.

But also notice how the `sign` is changing in front each time?

Answer 8

It's negative on the `even` derivatives,
and positive on the `odd` derivatives.

Answer 9

so how would you incorporate that

Answer 10

\[\Large g^{(n)}(x)\quad=\quad \frac{(-1)^n(n-x)}{e^x}\]So we want a (-1) to change between 1 and -1 each time like this.
What should our exponent on the -1 be though?
For n=1 do we want it to be -1 like it's set up now? :o\[\Large (-1)^1\quad=\quad -1\]\[\Large (-1)^2\quad=\quad +1\]

Answer 11

oh ok thanks

Answer 12

No the exponent is wrong :) fix it hehe

Answer 13

Oh actually.. this might be simpler...

Answer 14

If we factor a - out of each derivative... we can write them as,\[\Large g'(x)\quad=\quad \frac{-(x-1)}{e^x}\]\[\Large g''(x)\quad=\quad \frac{+(x-2)}{e^x}\]

Now the `odd` derivatives are `negative` so we can leave our (-1) to the n power without any trouble.

Answer 15

\[\Large g^{(n)}(x)\quad=\quad \frac{(-1)^n(x-n)}{e^x}\]

Answer 16

ok thanks