Derivative of function $v(t)= gt + v_0$ In this example, the derivative of $x$ is $1(x^{1-1})$, and this is based on the application of the power rule.
But, what if $x$ is $0$? Can $x$ be differentiated? Zero to the power of zero is undefined.

And therefore when the subsequent example shown below:
$s'(t) = v(t) = gt + v_0$, where I assume the derivative of the sum of $2$ functions was applied.
I don't quite understand how the derivative of $\frac{d}{dx}[v_0t]$ was derived as $v_0$, based on the previous example of the power rule.
Can I see $\frac{d}{dx}[v_0t]$ as $v_0$ as a constant, multiplied by the function $f(t)$, and I shall apply the power rule as $v_0 \times d/dx[t]$, and it becomes $v_0 \times \frac{d}{dx}[t] = v_0 *\times 1 $?
$\frac{d}{dx}[\frac{1}{2}gt^2+v_0t+s_0] $ is 


*

*$\frac{d}{dx}[\frac{1}{2} gt^2] = \frac{1}{2} \times 2 + gt^{(2-1) }= gt $

*$\frac{ d}{dx}[v_0t] = 1[v_0\times t^{(1-1)}]$

*$\frac{d}{dx}[s_0] = 0$, since the derivative of a constant is $0$



 A: If you have any doubt you can use the definition of derivative of a function at a point:
$$f'(c)=\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}\tag1$$
If $f(x)=x$ then
$$f'(0)=\lim_{h\to 0}\frac{0+h-0}{h}=\lim_{h\to 0}1=1$$
Explanation: the "power rule" and any other rule about how to obtain derivatives of functions are a consequence of $(1)$. 
So if you dont know if a rule can be applied in some context you can use the original definition of derivative at a point to know the derivative of a function at some problematic point.
A: Something you are confusing with is first plug then differentiate or first differentiate then plug
The rule often used is first differentiate then plug in the value 
So , in your example you should first differentiate and then plug 0 in function
Also you have written $\frac{d}{dx}$ instead of $\frac{d}{dt}$
This is a function of variable t not of x so you should differentiate with respect to t , if you do it wrt x , then the answer is 1 , which is not expected and not right too
If I understood your problem completely , then probably this would help.....!!!!
