3

Let $f$ and $g$ be $n$ times differentiable functions then prove that $$\frac {d^n}{dx^n}(fg)=\sum _{i=0}^n \binom{n}{ i} f^{(i)}g^{(n-i)} $$ where $f^{(k)} $ is $k$-th derivative with respect to $x $.

Now I started with Mathematical Induction. I know its true for $n=1$ so skipped it. Let it be true for $m<n $ thus $$\frac {d^m}{dx^m}(fg)=\sum_{i=0}^{m} \binom {m}{i}f^{(m)}g^{(m-i)} =s .$$ We need to prove this for $m+1$ . Note that $m+1 <n $ is also true. So we see that $$\frac {d^{m+1}}{dx^{m+1}}(fg)=\frac {d}{dx}(s).$$ But now problem here is that I don't know how to differentiate this whole $s$ as we cant use Leibniz rule to prove Leibniz rule.

Robert Z
  • 145,942
Archis Welankar
  • 15,910

2 Answers2

8

Since $$\frac {d^{m+1}}{dx^{m+1}}(fg)=\frac {d^{m}}{dx^{m}}\left(\frac {d}{dx}(fg)\right)=\frac {d^{m}}{dx^{m}}(f'g)+\frac {d^{m}}{dx^{m}}(fg')$$ apply Leibniz rule of order $m$ to the products $f'g$ and $fg'$. Hence \begin{align*} \frac {d^{m+1}}{dx^{m+1}}(fg)&=\sum_{i=0}^{m} \binom {m}{i}f^{(i+1)}g^{(m-i)}+\sum_{i=0}^{m} \binom {m}{i}f^{(i)}g^{(m+1-i)}\\ &=f^{(m+1)}g+ \sum_{i=1}^{m} \binom {m}{i-1}f^{(i)}g^{(m+1-i)} +\sum_{i=1}^{m} \binom {m}{i}f^{(i)}g^{(m+1-i)}+fg^{(m+1)}\\ &=f^{(m+1)}g+ \sum_{i=1}^{m} \left(\binom {m}{i-1}+\binom {m}{i}\right) f^{(i)}g^{(m+1-i)}+fg^{(m+1)}\\ &= \sum_{i=0}^{m+1} \binom {m+1}{i} f^{(i)}g^{(m+1-i)}. \end{align*} where we shifted the index $i$ in the first sum.

Robert Z
  • 145,942
3

I would go a different way than Robert Z. does.

If $\frac {d^n}{dx^n}(fg) =\sum _{i=0}^n \binom{n}{ i} f^{(i)}g^{(n-i)} $ then

$\begin{array}\\ \frac {d^{n+1}}{dx^{n+1}}(fg) &=\left(\frac {d^n}{dx^n}(fg)\right)'\\ &=\left(\sum _{i=0}^n \binom{n}{ i} f^{(i)}g^{(n-i)}\right)'\\ &=\sum _{i=0}^n \binom{n}{ i} \left(f^{(i)}g^{(n-i)}\right)'\\ &=\sum _{i=0}^n \binom{n}{ i} \left(f^{(i+1)}g^{(n-i)}+f^{(i)}g^{(n-i+1)}\right)\\ &=\sum _{i=0}^n \binom{n}{ i} f^{(i+1)}g^{(n-i)}+\sum _{i=0}^n \binom{n}{ i}f^{(i)}g^{(n-i+1)}\\ &=\sum _{i=1}^{n+1} \binom{n}{ i-1} f^{(i)}g^{(n-i+1)}+\sum _{i=0}^n \binom{n}{ i}f^{(i)}g^{(n-i+1)}\\ &=\sum _{i=1}^{n} \binom{n}{ i-1} f^{(i)}g^{(n-i+1)}+\binom{n}{ n} f^{(n+1)}g^{(0)}+\sum _{i=1}^n \binom{n}{ i}f^{(i)}g^{(n-i+1)}+\binom{n}{ 0}f^{(0)}g^{(n+1)}\\ &=f^{(n+1)}g+fg^{(n+1)}+\sum _{i=1}^{n} f^{(i)}g^{(n-i+1)}(\binom{n}{ i-1}+\binom{n}{ i})\\ &=f^{(n+1)}g+fg^{(n+1)}+\sum _{i=1}^{n} f^{(i)}g^{(n-i+1)}\binom{n+1}{ i})\\ &=\sum _{i=0}^{n+1} f^{(i)}g^{(n-i+1)}\binom{n+1}{ i})\\ \end{array} $

Either way, the derivative of a product is used.

marty cohen
  • 107,799