Derivative and integration I have 2 questions here that I have doubts with.

  
*
  
*If $y= \sin x $ , and $ y^{(n)}$ means "the $n$ derivative of $y$ with respect to $x$ " then the smallest positive integer $n$ for which $y^{(n)} = y $  is .... 
  

First I do the first derivative and 2nd and so on...
$y' = \cos x$ 
$y''= -\sin x $ 
does this mean $n= 2$ ? 
Next question is:


  
*If $f$ and $g$ are continuous functions , and if $f(x)\geq$, $\forall x\in\mathbb{R}$ , which of the following must be true? 
  


Am I right to say I And II ? 
Thanks ! 
 A: In regards to question 1, $\sin x \ne - \sin x$, because this implies $1 = -1$, which is certainly not true. To find the answer keep differentiating,
\begin{align*}
y' &= \cos x \\
y'' &= -\sin x \\
y^{(3)} &= -\cos x \\
y^{(4)} &= \sin x
\end{align*}
So $n=4$
In regards to question 2, II is a basic property of integrals, where I and III are fairly easy to find counter examples for, try $f(x) = e^x$ and $g(x) = x^3 - \sqrt{x}$
A: For question 1:
Your answer is wrong, because $-\sin x$ is not the same as $\sin x$, so $n$ has to be bigger than $2$.
Hints for question 2:
For I. and III. it is very easy to find counterexamples, just try some simple functions that come to your mind.
A: Question 1: No, $n \ne 2$, because $\sin x \ne -\sin x$. If you differentiate two more times, you will get
$$y'''=-\cos x$$
and
$$y''''=\sin x$$
So, since the fourth derivative is equal to $\sin x$, which is the same as the starting function, $n=4$.
Question 2: II is definitely always true because it is a basic property of definite integrals. I and III can be disproven easily using a few examples. Both can be disproven by letting $f$ and $g$ be constant functions.
A: For $n=2$ you get $-sin(x)$, which is different from $y=sin(x)$. Try $n=4$.
Second question, of the options you mention, option I is not in general true, for example $$\int_a^b x^2 \mathrm{d}x  = \int_a^b x*x \mathrm{d}x = \frac{1}{3}[b^3-a^3]$$ does not equal $$(\int_a^b x \mathrm{d}x) (\int_a^b x \mathrm{d}x) = \frac{1}{2}[b^2-a^2]*\frac{1}{2}[b^2-a^2]$$
A: Question 1
If $y = sinx$, then $y'''' = sinx$
If $y = cosx$, then $y'''' = cosx$
