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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.

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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 $0$

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N. F. Taussig
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ilovetolearn
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2 Answers2

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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.

Masacroso
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  • First, I apply the sum rule to break down the equation. Second, I apply the power rule and lastly the constant rule. – ilovetolearn Jun 25 '17 at 13:02
  • I don't see how can I apply the original derivative rule to this problem. Can you elaborate further? – ilovetolearn Jun 25 '17 at 13:02
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    I dont understand exactly what you are asking in the original question, but observe that if $v(t)=v_0\cdot t$ then $v'(t)=\frac{\mathrm d}{\mathrm d t}v(t)=v_0\cdot\frac{\mathrm d}{\mathrm dt}t=v_0\cdot 1$, because the derivative of $v(t)=t$ (respect to $t$, of course) is $1$. You asked the case when $t=0$, and this is was I had assumed it was your main question. – Masacroso Jun 25 '17 at 13:34
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    you have a typo in your original question, as other users said, you must write $\frac{\mathrm d}{\mathrm dt}$ instead of $\frac{\mathrm d}{\mathrm dx}$, otherwise the derivative of $v(t)$ respect to $x$ is zero (because $x$ is not a variable of $v$, so $v(t)$ is constant respect to $x$). – Masacroso Jun 25 '17 at 13:36
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    sorry about the ambiguity. I was trying to regurgitate the steps/calculations on how to derive the equations to make sure I understand the problem correctly. – ilovetolearn Jun 25 '17 at 14:21
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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.....!!!!

Atul Mishra
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