Is the function $f(x)=1^x=1$ considered an exponential function? I am confused about the following:

The exponential function (by definition) is a function of the form $f(x)=a^x$ where $a&gt;0$. However, when $a=1$, we get the constant function $f(x)=1^x=1$. Is the constant function $f(x)=1^x=1$ still considered an exponential function even though it does not have behave like an exponential function? Is the definition of the exponential function that I gave above (that I read in many textbooks) not entirely correct? Should we define the exponential function by:"a function of the form $f(x)=a^x$ where $a&gt;0$ and $a\neq 1$"? I welcome any answer. Thanks!
According to the definition you give, $f(x) = 1$ is indeed an exponential function. As you point out, it is indeed of the form $a^x$ for some $a&gt;0$. Wikipedia's definition explicitly rules out a base-1 exponential as being an exponential function, and this is understandable, as it is somewhat of a degenerate case if you consider it with other exponential functions. But it does satisfy the properties $f'(x) = k f(x)$ for some $k\in\mathbb{R}$ and $f(x+y) = f(x)f(y)$, so from those perspectives, it is also natural to include $f(x) = 1$ as an exponential function.
At the end of the day, it's a matter of convention what you decide to call an exponential function.
Well, an exponential function is one whose rate of change is proportional to its current $y$-value. i.e. one satisfying the differential equation $\frac{dy}{dx}=ky$ for some constant $k$.

$\frac{d}{dx}[a^x]=a^x\ln(a)=ka^x$ for $a \neq 0$.

Now, if $a=0$, i.e. if we want to evaluate $\frac{d}{dx}[1^x]$, we see that $\frac{d}{dx}[1^x]=1^x\ln(1)=1^x(0)=0 \neq k(1^x)$ (unless $k=0$, in which case $ \frac{dy}{dx}$ is not proportional to $y$).

So, essentially, your way is the way to go! As far as definition is concerned, an exponential function is any $f(x)=a^x$, where $a\in \Bbb{R^+} \setminus \{1\}$.