I am wondering whether or not this is the right sketch of the derivative in the picture below:
UPDATE:
Here is the function as it appears in the question (please disregard the pencil marks):
[Math] sketching derivative of a graph
derivativesgraphing-functions
Related Solutions
For any $x\in [0,2\pi]$ you get $f'(x)=\cos (x) -\sin (x)$ and $f''(x)=-\sin (x)-\cos (x)$. Therefore $$\begin{align} f'(x)=0&\iff \cos (x) -\sin(x)=0\\ &\iff \cos (x) =\sin (x)\\ &\iff x\in \{\frac{\pi}{4}, \frac{5\pi}{4}\} \end{align}$$
Which gives you the critical points $\displaystyle \frac{\pi}{4}$ and $\displaystyle \frac{5\pi}{4}$.
The last equivalence is easy to see geometrically if you look at the unit circle.
I don't understand why you're messing with $\cos ^{-1}$. Care to explain so we can help you?
Similarly, for $f''$ you'll get the zeros $\displaystyle \frac{3\pi}{4}$ and $\displaystyle \frac{7\pi}{4}$.
Some key points:
Zeroes
Local maxima/minima
Boundary/ies of the domain, if the domain is bounded from at least one direction
Boundary/ies of the support (where the function is non-zero), if the support is bounded in at least one direction
Asymptotes, if such exist
Points of inflection
Application:
$$y=\frac{1-x-x^2}{x^2}.$$
First, let's look for zeroes: $1-x-x^2=0$ when $$x=\frac{-1\pm \sqrt 5}2.$$
So estimate $\sqrt 5$ and use that, perhaps noting the exact values.
Next, we can rewrite this as $$y=\frac{1-x}{x^2}-1.$$ As $x$ increases or decreases without bound, this whole thing approaches $y=-1$, so draw that line.
$$y'=\frac{-x^2+2x(x-1)}{x^4}=\frac{x-2}{x^3},$$ which is $0$ exactly at $2$.
$$y''=\frac{x^3-3x^2(x-2)}{x^6}=\frac{-2x+6}{x^4},$$ which is positive at $2$, so $x=2$ is a local minimum. Draw that.
The function is undefined at $x=0$ and in fact has no limit. Figure out which ways it goes.
Best Answer
I recommend that you position a set of axes directly below the given graph. You then draw the graph of the derivative keeping in mind that the slope of the original function tells you the value of the derivative function. Specifically, the marked points at $(-2,-5)$ and $(1,25)$ are points where the slope is zero. (Whether the slope is genuinely zero at $(-2,-5)$ is debatable but I believe that is the likely intent.) You can project down to the $x$-axis on your new axes to obtains roots of the derivative. However you draw the derivative, you must go through those two points. Furthermore, you cannot cross at any other point. I guess that gets you to here:
Now, you need to draw a graph that goes through those two dots and stays on the positive or negative side of the $x$-axis as indicated by the slope. I guess the simplest graph that does that looks like so: