\displaystyle{2}+{\cos{{x}}} Explanation: The derivative of this function is equal to the sum of the derivatives of each term. That is, the derivative equals \displaystyle\frac{{d}}{{\left.{d}{x}\right.}}{\left[{2}{x}\right]}+\frac{{d}}{{\left.{d}{x}\right.}}{\left[{\sin{{x}}}\right]} ...

This is an interesting question, because you wouldn't necessarily thing that the sum of two sine waves would give another sine wave and not some other wavy thing. The key to solving this problem is ...

\displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\sin{{\left({4}\pi{x}\right)}}}\right)}={4}\pi{\cos{{\left({4}\pi{x}\right)}}} Explanation: Using the chain rule: \displaystyle\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({\sin{{\left({4}\pi{x}\right)}}}\right)}=\frac{{{d}{\sin{{\left({4}\pi{x}\right)}}}}}{{{d}{\left({4}\pi{x}\right)}}}\frac{{{d}{\left({4}\pi{x}\right)}}}{{{\left.{d}{x}\right.}}}={\cos{{\left({4}\pi{x}\right)}}}\cdot{4}\pi

\displaystyle{x}=\frac{{{n}\pi}}{{2}}+\frac{\pi}{{8}} , where \displaystyle{n} is an integer Explanation: As \displaystyle{\sin{{4}}}{x}={1}={\sin{{\left(\frac{\pi}{{2}}\right)}}} and ...

The Mean Value Theorem states: If f(x) is continuous on [a,b] and differentiable on (a,b), then there exists at least on c\in (a,b) such that f'(c) = \frac{f(b)-f(a)}{2\pi}. Our function ...

As sketched in the comments: This follows from the Intermediate Value Theorem applied to the function g(x)=sin(x)-\frac x2 Indeed, g(\frac {\pi}2)=1 - \frac {\pi}4>0 while g(\pi)=-\frac {\pi}2<0 ...