find the higher order derivative:

Question

d^n/dx^n (2^x)

Answer 1

Hassan's answer is correct, but one line of his work, namely

d^2/dx^2 (2x)=(ln2)² 2², 

contains an error--last factor should be 2^x.

Answer 2

First start with first derivative of y=2^x

rewrite 2^x as e^xln2

so d/dx (2^x)=ln2 2^x

then next do second derivative

d^2/dx^2 (2x)=(ln2)² 2²

you continue in this manner with other order of derivative

you can see that

d^n/dx^x(2^x)=(ln2)ⁿ 2^x