Induction: prove that f^n(x) = (-1)^n e^(-x)(x - n) for all

[stars584]

Can someone please guide me through this problem. I know that I have to take the derivative of (-1)^n e^(-x) (x-n).

Derivative= n(-1)^(n-1) e ^(-x) ???

Let f(x)= xe^(-x). Prove that f^(n) (x) = (-1)^n e^(-x) (x - n) for every positive integer n.

Proof by Induction:

Base Case: When n = 1, the statement is (-1)^1 e^(-x) (x-1)=

 

[stapel]

For n = 1, you're taking the first derivative. Does the first derivative fit the pattern? If so, fill that in for the "n = 1" case.

For n = k, write out the assumption.

For n = k + 1, write out the left-hand side, note that f^k+1(x) = d[f^k(x)]/dx, and differentiate the assumption step. Does this also fulfill the required pattern?

If you get stuck, please reply showing how far you have gotten. Thank you.

Eliz.

 

[stars584]

BC. when n=1, the stmt is (-1)^1 e^(-x)(x-1)= -

f'(x)= n(-1)^(n-1) (-e^(-x)) (1-n) right???

so if above derivative is correct than for n=1; 1(-1)^(1-1) (-e^(-x)) (1-1)
(-1^0) (-e^(-x)) (0)
-1 (-e^(-x)) (0)

I am going in the right direction?

 

[stapel]

BC. when n=1, the stmt is (-1)^1 e^(-x)(x-1)= -

I'm sorry, but I don't know what this means...? (Using standard English might be helpful.)

When n = 1, you are taking the first derivative of f(x) = x e^-x. What is that derivative?

According to the formula you want to prove, the derivative should equal (1)(-1)^1-1 e^-x (x - 1) = x e^-x - e^-x. Is this what you get? That is, can you show that the formula works for n = 1?

And so forth....

Eliz.