What is an efficient method to find the fifth derivative of $e^{kx}x^{-1}$?

Question

What is the fifth derivative of $$e^{kx}x^{-1}$$ and how can it be computed efficiently?

Obviously, this can be computed using successive applications of the quotient rule, but the calculations become unmanageable. Moreover, there are clear patterns that should allow us to "collapse" most of those calculations and deduce a cleaner method. What are they?

Background: This derivative comes up repeatedly in What is the most efficient method to evaluate this indefinite integral? , yet none of the answers show how to calculate it.

Work so far: I attempted to solve this using operator notation: $$\begin{align*} D^5e^{kx}x^{-1} &= k^5e^{kx}(D+k)^5x^{-1} \\ &= k^5e^{kx}(D^5 + 5kD^4 + 10k^2D^3 + 10k^3D^2 + 5k^4D + k^5)x^{-1} \end{align*}$$ but couldn't keep things neat or well-patterned for the next step (which begins $k^5e^{kx}(-5!x^{-6} + 4!5x^{-5} -3!10x^{-4} + ...)$ which does not seem to agree with e.g. this reference or Wolfram Alpha). Is my approach correct? If yes, how do I make the calculations manageable? If not, why not?

Answer 1

Let $g(x)=\frac{e^x}x.$ Then $f(x)=kg(kx),$ so,$$f^{(5)}(x)=k^6 g^{(5)}(kx).$$

So you only need to solve the case $k=1.$

Now, if $$g^{(j)}(x)=h_j(x)e^x$$ then $$g^{(j+1)}(x)=\left(h_j(x)+h_j'(x)\right)e^x$$

So $h_{j+1}(x)=h_j(x)+h_j'(x),$ with $h_0=\frac1x.$

Then $h_1(x)=\frac1x-\frac1{x^2},$ $h_2(x)=\frac1x-2\frac1{x^2}+\frac2{x^3}.$ In general, you get $$h_j(x)=\sum_{i=0}^j(-1)^i\binom ji\frac{i!}{x^{i+1}}$$

Answer 2

There is a general formula due to Leibniz, which can be proved easily by induction $$ (fg)^{(n)}= \sum_{j=0}^n{n\choose j}f^{(n-j)}g^{(j)}$$ with convention $h^{(0)}=h.$ For $f(x)=e^{kx}$ and $g(x)=x^{-1}$ we get $$ (e^{kx}x^{-1})^{(n)}=\sum_{j=0}^n{n\choose j} k^{n-j}e^{kx}(-1)^j j!\,x^{-j-1}\\ =k^nn!\,e^{kx}\sum_{j=0}^n{(-1)^j\over (n-j)!}k^{-j}x^{-j-1}$$ We use the convention $(-1)^0=1.$

Answer 3

Yes.

$$D^n(e^{kx}x^{-1})= e^{kx}x^{-(n+1)} \left(\sum_{i=0}^{n} (-1)^{i}i!{n \choose i}k^{n-i}x^{n-i}\right) $$

It is the "simplier" I think we can get.

It uses Leibniz.

Answer 4

Not pretty, but here is one approach using Taylor series:

With $f(x) = e^{kx} {1 \over x}$ we have $f(x+h) = f(x) {x \over x+h} e^{kh} = \sum_{i=0}^\infty (-1)^i ({h \over x})^i \sum_{j=0}^\infty {(kh)^j \over j!}$. The Taylor series is $f(x+h) = \sum_{i=0}^\infty {1 \over i!}f^{(i)}(x) h^i$, so $f^{(5)}(x) = 5! f(x) \sum_{l=0}^5 (-1)^l{1 \over x^l} {k^{5-l} \over (5-l)!}$.

Answer 5

Yet another approach. Let $y(x,k) = \frac{\partial^t}{\partial x^t} e^{k x}x^{-1}$. Then note that $$ \frac{\partial y}{\partial k} = \frac{\partial^t}{\partial x^t} e^{k x} = k^t e^{k x}.$$ This leads to an easy integration by parts formula which is identical to Thomas Andrew's post.