# Help needed Thank You in Advance

• May 14th 2009, 01:24 PM
Rylude
Help needed Thank You in Advance
Question: Using binomial coefficients, derive, a formula for the Nth derivative of the product of two functions.

Thank You

Dave
• May 15th 2009, 12:00 AM
Spec
$y(x)=f(x)g(x)$
$y'(x)=f'(x)g(x)+f(x)g'(x)$
$y''(x)=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)=$ $\left(\begin{array}{c}2\\0\end{array}\right)f''(x) g(x)+\left(\begin{array}{c}2\\1\end{array}\right)f '(x)g'(x)+\left(\begin{array}{c}2\\2\end{array}\ri ght)f(x)g''(x)$
$y^{(3)}(x)=f^{(3)}(x)g(x)+3f''(x)g'(x)+3f'(x)g''(x )+f(x)g^{(3)}(x)=$ $\left(\begin{array}{c}3\\0\end{array}\right)f^{(3) }(x)g(x)+\left(\begin{array}{c}3\\1\end{array}\rig ht)f''(x)g'(x)+\left(\begin{array}{c}3\\3\end{arra y}\right)f'(x)g''(x)+\left(\begin{array}{c}2\\0\en d{array}\right)f(x)g^{(3)}(x)$

Can you continue?

Leibniz rule (generalized product rule) - Wikipedia, the free encyclopedia
• May 15th 2009, 03:17 AM
HallsofIvy
You can prove the general formula by induction:

$(f^{n+1}) ' = (f^{n} f) ' = (f^n) ' f + f^n f '$.
• May 19th 2009, 08:20 AM
chase2910
sorry i dont understand how to get the formula after that, i understand the use of the product rule and the binomial thoerem, but i dont understand wer to go after that