How would you differentiate something like:
$\displaystyle y = 2^x$ With respect to $\displaystyle x$?
Hello,
Yes it is.
Actually, you can see it this way :
$\displaystyle a^{f(x)}=e^{f(x) \ln a}$
--> $\displaystyle \left(a^{f(x)}\right)'=\ln a \cdot f'(x) e^{f(x) \ln a}=f'(x) \ln a \cdot a^{f(x)}$
Here, $\displaystyle f(x)=x-1$
You could also find this result by applying the chain rule and using wingless's formula
In another way, you can see that $\displaystyle 2^{x-1}=\frac 12 \cdot 2^x$