What is the derivative of zero and why is it a special function?

[PrakashPhy]

Does the derivative of zero equal to zero make any sense?

 

[Nick R]

That the derivative of 0 is 0 means that zero doesn't vary at all when some independent variable is varied.

edit: actually I guess you'd need to know that all derivatives (second, third and so on) of 0 are 0 to say that. But yeah any order derivative of a constant wrt to any variable is 0 - that's what makes it a constant.

 

[JJacquelin]

Question : Does the derivative of a constant equal to zero make any sense?
Answer : Yes, it makes sens.
Question : Does zero is a constant ?
Answer : Yes, it is.
So, then ...

 

[HallsofIvy]

\frac{df}{dx}= \lim_{h\to 0}\frac{f(x+h)- f(x)}{h}

If f(x)= 0 for all x, then f(x+h)= 0 and this becomes
\frac{df}{dx}= \lim_{h\to 0}\frac{0- 0}{h}= \lim_{h\to 0} 0= 0.

In fact, exactly the same proof shows that the derivative of any constant function is 0. That's usually one of the first things one learns about the derivative.

Another, perhaps even more fundamental proof: The derivative of a function, at a given value of x, is the slope of the tangent line to the graph of thefunction at that point. In this case the graph of f(x)= 0 (or any constant) is a horizontal straight line so the tangent line to the graph is the graph and has slope 0.

 

[g_edgar]

The derivative of the constant function with value 0 is again that constant function. Perhaps that is what you mean. If f is a function and f(a)=0 for a particular a, then it does not follow that f'(a)=0, however.

 

[HallsofIvy]

g_edgar makes a good point, PrakashPhy. Each person's response here has assumed you were talking about the function f(x)= 0 that is 0 for all x. You only differentiate functions, not numbers.

If you are talking about some function, f(x), such that f(x)= 0 for a specific x, then its derivative at that point could be anything. The derivative of a function, at a given x, is independent of its value at that point.

 

[mosimane]

if you where to think of 0 as a funtion of f that would implie that for every values of x in the domain of f the output parameter will be 0 then f is continuos then f is differantible thus f(x)=0 then f'(x)=0

 

[Tac-Tics]

Like some people hinted at above, it depends on what you mean.

The derivative is an operation on a function, not an operation on a number.

However, sometimes, when we say something like "the derivative of 0" we actually mean "the derivative of f where f(x) = 0".

Constant functions, such as f(x) = 0, always have the constant zero function for a derivative: f'(x) = 0.

 

[Finbar]

Actually if we take the derivative of f(x) we have f'(x).

But if f(x_0)=0 it is not generally true that f'(x_0)=0.

for example f(x)=x

f(0)=0

f'(0)=1

So its not generally true that the derivative of a function is zero where the function is zero.

Of coarse if f(x) = 0 for all x then f'(x)=0.

 

[PrakashPhy]

Thank you all.
I had a mistake in understanding what derivative actually is. I thought of taking derivative of some numbers, where i had mistake. I should take derivative of functions only.
With these i come to a conclusion:
" derivative of a function f(x)=0 makes a sense, because for every x in cartisean plane the value of y is zero so it gives a straight line concident with x axis, the tangent on every point on which make angle of zero radian with x-axis so the derivative ( slope is zero )"

Please suggest me if i have the wrong understanding.

Thank you all for your support again.

 

[HallsofIvy]

Thank you all.
I had a mistake in understanding what derivative actually is. I thought of taking derivative of some numbers, where i had mistake. I should take derivative of functions only.
With these i come to a conclusion:
" derivative of a function f(x)=0 makes a sense, because for every x in cartisean plane the value of y is zero so it gives a straight line concident with x axis, the tangent on every point on which make angle of zero radian with x-axis so the derivative ( slope is zero )"

Please suggest me if i have the wrong understanding.

Thank you all for your support again.

That's correct and is, in fact, what I said in response #4:
"Another, perhaps even more fundamental proof: The derivative of a function, at a given value of x, is the slope of the tangent line to the graph of thefunction at that point. In this case the graph of f(x)= 0 (or any constant) is a horizontal straight line so the tangent line to the graph is the graph and has slope 0. "

 

[Tac-Tics]

Thank you all.
I had a mistake in understanding what derivative actually is. I thought of taking derivative of some numbers, where i had mistake. I should take derivative of functions only.

The derivative is the first operator you learn in math that operates on functions instead of numbers (but the notation for it doesn't make this clear!)

If real numbers are designated R, then 0, 1, pi, e, etc, are all elements of R.

If functions from X to Y are designated X->Y, then our familiar case of real functions can be notated as R->R. The set R->R includes sine, polynomials, constant functions, piece-wise defined functions, etc.

The derivative is *also* a function, but a more special one :) It takes real functions as input and gives you back another real function. The derivative belongs to (R->R)->(R->R). Other "functions" in (R->R)->(R->R) include the anti-derivative operation.

The X->Y notation is useful (especially in type theory and category theory) because it tells you can and cannot use as input to a function.