I know that the derivative is 0 for all x but can somebody explain this to me. I understand derivatives but I have been having a lot of trouble here. I know that $ceil(x)=(x+1/2) – (arctan(tan(pi*(x+1/2))))/(pi)$ for all non integer x so I differentiated that on wolfram alpha https://www.wolframalpha.com/input/?i=derivative+of+(x%2B1%2F2)+-+(arctan(tan(pi(x%2B1%2F2))))%2Fpi and got $1-(csc^2(pi*x)/(cot^2(pi*x)+1)$ and using the desmos graphing calculator to graph this I got https://www.desmos.com/calculator/mmzdqg6nba which is 0 for all non-integer x. can anybody show me an easier way to solve this problem.
Derivatives – Proof and explanation of the derivative of ceil(x)
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Q1. It means exactly what it says. :-) How much does one variable change, with respect to (that is, in comparison to) another variable? For instance, if $y = 3x$, then the derivative of $y$, with respect to $x$, is $3$, because for every unit change in $x$, you get a three-unit change in $y$.
Of course, that's not at all complicated, because the function is linear. With a quadratic equation, such as $y = x^2+1$, the derivative changes, because the function is curved, and its slope changes. Its derivative is, in fact, $2x$. That means that at $x = 1$, an infinitesimally small unit change in $x$ gives a $2x = 2$ unit change in $y$. This ratio is only exact right at $x = 1$; for example, at $x = 2$, the ratio is $2x = 4$.
This expression is the limit of the ratio $\frac{\Delta y}{\Delta x}$, the change in $y$ over the change in $x$, over a small but positive interval. The limit as that interval shrinks to zero is $\frac{dy}{dx}$.
Q2. You will rarely see, at this stage, $\frac{d}{dx}$ by itself. It will be a unary prefix operator, operating on an expression such as $x^2+1$. For instance, we might write
$$ \frac{d}{dx} \left(x^2+1\right) = 2x $$
It just means the derivative of the expression that follows.
Q3. This is an unusual formulation. Ostensibly, though, it would mean the derivative of the operand with respect to $f(x)$, which you can obtain using the chain rule:
$$ \frac{dx}{df(x)} = \frac{\frac{dx}{dx}}{\frac{df(x)}{dx}} = \frac{1}{f'(x)} $$
and
$$ \frac{d}{df(x)} g(x) = \frac{\frac{dg(x)}{dx}}{\frac{df(x)}{dx}} = \frac{g'(x)}{f'(x)} $$
The derivative of a function gives the slope of the tangent line to that function at the same $x$ co-ordinate, so looking at a graph of the derivative tells you something about how the function is moving at various points. In particular:
If the derivative is 0, then the tangent line is horizontal, so the function must go completely flat at that point (either a local maximum or minimum, or a point of inflexion). For example, on your graph, the derivative is zero at -1, 0 and 1, which are all points where the function itself is completely flat.
If the derivative is positive, then the tangent line is diagonal going up-right, so the function is increasing at that point. For example, the derivative in your graph is positive between -1 and 0, and from 1 onwards, and that's where the graph slopes up.
If the derivative is negative, then the tangent line is diagonal down-right, so the function is decreasing at that point.
If you want to get fancier, then since bigger values of the derivative (either positive or negative) mean more significant slopes, then the fact that your derivative trends to infinity in both directions means that your graph will get more extremely sloping as it goes along. By comparison, if the derivative stayed very small, then the function would be much more gentle even as it trended wherever it was going.
Best Answer
The ceiling function is piecewise constant and the derivative of a constant is zero.
The ceiling function is discontinuous at integer values and is non-differentiable there.