Question

Prove Leibniz Formular:product rule

Answer 1

\[(fg)^{(n)}=\sum_{k=0}^{n}\left(\begin{matrix}n \\ k\end{matrix}\right)f^{k}g^{n-k}\]
where\[f^{(n)}=\frac{ d^{(n)} f}{ dx^{n} }\]

Answer 2

link can also help

Answer 3

I am stuck on \[\left(\begin{matrix}m \\ k-1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=?\]

Answer 4

http://physicspages.com/2011/03/22/generalized-product-rule-leibnizs-formula/

Answer 5

http://www.math.ucsd.edu/~wgarner/math20a/prodrule.htm

Answer 6

\[\left(\begin{matrix}m \\ k-1\end{matrix}\right)+\left(\begin{matrix}m \\ k\end{matrix}\right)=\frac{ m! }{ (k-1)!(m-k+1)! }+\frac{ m! }{ k!(m-k!) }\]\[=\frac{ m! }{ (k-1)!(m-k)! }\left[ \frac{ 1 }{ m-k+1 }+\frac{ 1 }{ k } \right]\]\[=\frac{ m! }{ (k-1)!(m-k)! }\left[ \frac{ k+m-k+1 }{ k(m-k+1) } \right]\]\[=\frac{ m! }{ (k-1)!(m-k)! }.\frac{ m+1 }{ k(m-k+1) }\]\[=\frac{ (m+1)! }{ k!(m-k+1)! }=\left(\begin{matrix}m+1 \\ k\end{matrix}\right)\]

Answer 7

@Jonask hope this helped

Answer 8

thanks guys ,appreciated

Answer 9

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