For $f(x) = p(x)g(x)$,

By writing them out repeatedly, I am guessing that $f^{(k)}(x) = \sum_{i=0}^{k} {k \choose i} p^{(k-i)}(x) g^{(i)}(x)$ (where $f^{(k)}(x)$ : $k$-th derivative of $f(x)$). I tried proving this by mathematical induction, but am stuck on how to proceed from the induction hypothesis.

how can I prove by mathematical induction that

$f^{(k)}(x) = \sum_{i=0}^{k} {k \choose i} p^{(k-i)}(x) g^{(i)}(x)$ ?

Please help! Thank you in advance.


1 Answer 1


The base of induction is clear. For the the inductive step: $$f^{(k+1)}(x)=\frac{d}{dx}f^{(k)}(x) = \frac{d}{dx}\Bigg(\sum_{i=0}^{k} {k \choose i} p^{(k-i)}(x) g^{(i)}(x)\Bigg)=\sum_{i=0}^{k} {k \choose i}\cdot \frac{d}{dx}\bigg(p^{(k-i)}(x) g^{(i)}(x)\bigg)$$ $$=\sum_{i=0}^{k} {k \choose i} \bigg(p^{(k-i)}(x)g^{(i+1)}(x)+p^{(k-i+1)}(x)g^{(i)}(x)\bigg)$$ $$=\sum_{i=0}^{k+1} \bigg({k \choose i}+{k \choose {i-1}}\bigg) p^{(k+1-i)}(x) g^{(i)}(x)=\sum_{i=0}^{k+1} {{k+1} \choose i} p^{(k+1-i)}(x) g^{(i)}(x) $$ And the last move was due to the Pascal identity : $\binom{k}{i}+\binom{k}{i-1}=\binom{k+1}{i}$


Not the answer you're looking for? Browse other questions tagged .