Higher order or тth order derivative of an exponential function

Question

How should I approach the task of finding the 100th order derivative of a function $$y = (x+1)2^{x+1}$$ Before, I worked with nth order derivatives of trigonometric functions and applied Euler's formula to simplify the task. Here I am not sure where to start. Should I obtain the first, the second, the third and so on derivatives and look for a pattern or is there a more convenient way?

Answer 1

Let $f_c(x)=(x+c)2^{x+1}$. We can easily calculate the derivative of one of these functions, as $$f_c’(x)=\log(2)\left(\left(x+c+\frac{1}{\log(2)}\right)2^{x+1}\right)=\log(2)f_{c+\frac{1}{\log(2)}}(x).$$ By a simple induction, then, the 100-th derivative of $f_1(x)$ will be $$(\log(2))^{100}f_{1+\frac{100}{\log(2)}}(x)= (\log(2))^{100}\left(x+1+\frac{100}{\log(2)}\right)2^{x+1}.$$

Answer 2

The expression be split in two terms $(y=y_1+y_2)$ where $y_1=2xe^{xln2}$ and $y_2=2e^{xln2}$. The general pattern for derivatives are $y_1^{(n)}=2x(ln2)^ne^{xln2}+2n(ln2)^{n-1}e^{xln2}$ and $y_2^{(n)}=2(ln2)^ne^{xln2}$.

To summarize $y^{(n)}=2^{x+1}((x+1)(ln2)^n+n(ln2)^{n-1})$.