Question: Using binomial coefficients, derive, a formula for the Nth derivative of the product of two functions.
Thank You
Dave
$\displaystyle y(x)=f(x)g(x)$
$\displaystyle y'(x)=f'(x)g(x)+f(x)g'(x)$
$\displaystyle y''(x)=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)=$ $\displaystyle \left(\begin{array}{c}2\\0\end{array}\right)f''(x) g(x)+\left(\begin{array}{c}2\\1\end{array}\right)f '(x)g'(x)+\left(\begin{array}{c}2\\2\end{array}\ri ght)f(x)g''(x)$
$\displaystyle y^{(3)}(x)=f^{(3)}(x)g(x)+3f''(x)g'(x)+3f'(x)g''(x )+f(x)g^{(3)}(x)=$ $\displaystyle \left(\begin{array}{c}3\\0\end{array}\right)f^{(3) }(x)g(x)+\left(\begin{array}{c}3\\1\end{array}\rig ht)f''(x)g'(x)+\left(\begin{array}{c}3\\3\end{arra y}\right)f'(x)g''(x)+\left(\begin{array}{c}2\\0\en d{array}\right)f(x)g^{(3)}(x)$
Can you continue?
Leibniz rule (generalized product rule) - Wikipedia, the free encyclopedia