Proof by Mathematical Induction

We will prove that for the function f(x) = (2 + x)e^{-x} , the n^{\text{th}} derivative is given by:

f^{(n)}(x) = (-1)^n e^{-x} (x + 2 - n)

Base Case ( n = 1 ):

First, we verify the statement for n = 1 .

From (a) we know that:

f'(x) = (-x - 1)e^{-x}

According to the formula:

f^{(1)}(x) = (-1)^1 e^{-x} (x + 2 - 1) = -e^{-x}(x + 1) = (-x - 1)e^{-x}

Since both expressions for f'(x) match, the base case holds true .

Induction Hypothesis:

Assume that the statement holds for some arbitrary positive integer k . That is, assume:

f^{(k)}(x) = (-1)^k e^{-x} (x + 2 - k)

Inductive Step:

We need to show that the statement holds for n = k + 1 . That is, we need to prove:

f^{(k+1)}(x) = (-1)^{k+1} e^{-x} (x + 2 - (k + 1)) = (-1)^{k+1} e^{-x} (x + 1 - k)

Differentiate f^{(k)}(x) with respect to x :

f^{(k+1)}(x)=\frac{\differentialD}{\differentialD x}\left[(-1)^{k}e^{-x}(x+2-k)\right]

Applying the product rule:

f^{(k+1)}(x)=(-1)^{k}\cdot\frac{\differentialD}{\differentialD x}\left[e^{-x}\right]\cdot(x+2-k)+(-1)^{k}e^{-x}\cdot\frac{\differentialD}{\differentialD x}\left[x+2-k\right]

f^{(k+1)}(x) = (-1)^k (-e^{-x}) (x + 2 - k) + (-1)^k e^{-x} (1)

f^{(k+1)}(x) = -(-1)^k e^{-x} (x + 2 - k) + (-1)^k e^{-x}

f^{(k+1)}(x)=(-1)^{k+1}e^{-x}(x+2-k)-(-1)^{k+1}e^{-x}

Factor out (-1)^{k+1}e^{-x}:

f^{(k+1)}(x)=(-1)^{k+1}e^{-x}\left[(x+2-k)-1\right]

f^{(k+1)}(x)=(-1)^{k+1}e^{-x}\left[\left(x+1-k\right)\right]

This matches the required form:

f^{(k+1)}(x) = (-1)^{k+1} e^{-x} (x + 1 - k)

Conclusion:

We have verified the base case for n = 1 and shown that if the statement holds for n = k , then it holds for n = k + 1 . By the principle of mathematical induction, the statement

f^{(n)}(x) = (-1)^n e^{-x} (x + 2 - n)

holds for all n \in \mathbb{N} .

Proof and Reasoning