Question

Find a General formula for f^(n) (x)?

If f(x)= 3xe^x

Answer 1

`f^(n)(x) = 3n e^x + 3xe^x`, where `n` is an integer

Explanation:

We can immediately see that

`f'(x) = 3e^x + 3xe^x`
`f''(x) = 3e^x + 3e^x + 3xe^x = 6e^x + 3xe^x`
`f'''(x) = 6e^x +3e^x + 3xe^x = 9e^x + 3xe^x`

So the pattern is

`f^(n)(x) = 3n e^x + 3xe^x`, where `n` is an integer

Hopefully this helps!

Answer 2

`d^n/dx^n (3xe^x)=3(x+n)e^x`

This can be proved by a simple application of Leibnitz' rule of successive differentiation.

Explanation:

Leibnitz rule of successive differentiation of products (aka general Leibnitz rule ) is a generalization of the well known product rule (aka Leibnitz rule) :

`d/dx(uv) = u (dv)/dx+(du)/dx v`

to the `n`-th derivative:

`d^n/dx^n (uv) = ""^nC_0 u (d^nv)/(dx^n) + ""^nC_1 (du)/dx (d^(n-1)v)/dx^(n-1)+...+""^nC_n(d^n u)/(dx^n)v`
`qquad = sum_{r=0}^n ""^nC_r (d^ru)/(dx^r) (d^(n-r)v)/(dx^(n-r))`

Note that the formula for the `n`-th derivative of a product looks very similar to the binomial expansion! Now, in general, this formula has `n+1` terms - but if one of the factors of the product is a polynomial of a small degree `m (< n)`, we can choose it as the factor `u`, and exploit the fact that differentiating it more than `m` times will make it vanish.

In this case, we take `u=3x` and `v=e^x`. Since `(d^ru)/dx^r` vanishes for `r>=2`, we will have only two terms :

`d^n/dx^n (3xe^x) = 3x(d^n e^x)/(dx^n)+n d/dx(3x) d^(n-1)/dx^(n-1)(e^x)`
`qquad = 3xe^x+3n e^x = 3(x+n)e^x`

(we have used the fact that differentiation `e^x` with respect to `x` keeps it unchanged.)