Derivative of degree k for $f(t)$ $=$ $1 \over {1 + t}$

Question

Given

$f: \Bbb R \setminus \{-1\} \rightarrow \Bbb R$,

$f(t)$ $=$ $1 \over {1 + t}$,

I would like to calculate the derivative of degree $k$.

Approach

First, we try to examine if there is a certain pattern that can be observed.

Applying the quotient rule a few times, we receive:

$f'(t)$ $=$ $-1 \over {(1 + t)^2}$,

$f''(t)$ $=$ $2 \over {(1 + t)^3}$,

$f'''(t)$ $=$ $-6 \over {(1 + t)^4}$,

...

The structure of the derivatives becomes quite obvious. Therefore, we claim:

$f^k$ $=$ $(-1)^k$ $k! \over (1 + t)^{k+1}$

Proof by induction

For $k = 0$ we get

$(-1)^0$ $0! \over (1 + t)^{0 + 1}$ $=$ $1 \over (1 + t)$

Now, assume the statement has already been proven for $k = n.$ We have to show that it also holds for $k = n+1.$

$f^{(n+1)}$ $=$ $(-1)^{n+1}$ $(n+1)! \over (1 + t)^{n+2}$ $=$ $(-1)^n$ $n! \over (1 + t)^{n+1}$ $(-1)$ $n+1 \over (1 + t)$ $= f^n$ $(-1)$ $n+1 \over (1 + t)$ $=$ ...,

and this is where I'm stuck. I just have to show that the expression on the right side is the first derivative of the function, right? But this doesn't work out, I guess.

Answer 1

Assume the statement has been already proven for $k = n.$

That is $$f^{n}=(-1)^n\frac{n!}{(1+t)^{n+1}}$$

Now differentiate $f^n$ , then you will get

$$\frac{df^{n}}{dt}=(-1)^n \cdot n!\cdot \frac{d}{dt}\frac{1}{(1+t)^{n+1}}$$

$$\frac{df^{n}}{dt}=(-1)^n \cdot n!\cdot \frac{(-1)(n+1)}{(1+t)^{n+2}}$$

Thus $$f^{n+1}=(-1)^{n+1} \cdot (n+1)!\cdot \frac{1}{(1+t)^{n+2}}$$

Answer 2

If $$f^{(n)}(t)=(-1)^n\frac{n!}{(1+t)^{n+1}}$$ then $$f^{(n+1)}(t)=\left(f^{(n)}\right)'(t)=(-1)^nn^{-(n+2)}=\left(-1\right)^{n+1}\frac{(n+1)!}{(1+t)^{n+2}}$$

Answer 3

In the induction step, you assume that the formula is true for $k = n$. But when you are showing for $k = n+1$ you are saying that $f^{(n+1)}(t) = (-1)^{n+1} \frac{(n+1)!}{(1+t)^{n+2}}$, which is exactly what you want to prove! Your reasoning is circular.

What you know is that $f^{(n)}(t) = (-1)^n \frac{n!}{(1+t)^{n+1}}$. Then $f^{(n+1)}(t) = (f^{(n)}(t))' = \left( (-1)^n \frac{n!}{(1+t)^{n+1}} \right)'$. I leave the rest to you.

Answer 4

When you do the inductive step you have to start from the hypothesis that the statement holds for $k=n$ and then conclude that it holds for $k=n+1$. In your proof you already assume the conclusion. Instead suppose that the statement holds for $k=n$ and note that $$ \begin{align} f^{(n+1)}(t) &=\frac{d}{dt}\left( (-1)^n \frac{n!}{(1+t)^{n+1}}\right)\\ &=(-1)^n \times(1-n)\times n!(1+t)^{-(n+2)}\\ &=(-1)^{n+1} \frac{(n+1)!}{(1+t)^{n+2}} \end{align} $$ where we have used the power rule in the second line and the inductive hypothesis in the first line.