I was answering some questions for incoming quizzes when a wild question appears. It looks like this:

> 
> Suppose that $f$ is an odd function for all $x$. Suppose that $f'(x\_o)$ exists. Which of the following must be necessarily be equal to $f'(-x\_o)$?
> 
> 
> A. $f'(x\_o)$
> 
> 
> B. $\frac{1}{f'(x\_o)}$
> 
> 
> C. $-f'(x\_o)$
> 
> 
> D. $-\frac{1}{f'(x\_o)}$
> 
> 
> 

**My work**

I think that the answer would be **C**. Why? The funtion I know that fits in the description of the equation above would be the cosine function $y = \cos x$.

Differentiating $y = \cos(x)$, it becomes $y' = - \sin(x)$. Then everybody knows the property of sine function: $\sin (-x) = -\sin(x)$. 

So....

$y' = -\sin(x)$ is just equal to $y' = \sin(-x)$. So if the problem states that "**Which of the following must be necessarily be equal to $f'(-x\_o)$?**" , then, I think.....$f'(-x\_o)$ would be equal to $-f'(x\_o)$ (which is the choice **C**). It works on cosine function......

Is my logic correct?

Because it seems that the function being described could be anything other than cosine function...but I don't know what are other functions 
that could fit on the problem statement other than the cosine function.