let f(x)=x+(x^2)*sin(2/x) if x is not 0 , and f(0)=0
1)find the derivative of f at x=0
2) if a>0, then show that f is not increasing on (-a,a)
1) First note that the function is continuous at $\displaystyle x=0$, so a derivative could exist. To find the derivative for every point except $\displaystyle x=0$, just use the product and chain rule and note that
$\displaystyle \frac{d}{dx}sin(2/x)=\frac{-2}{x^2}\cos(2/x)$.
So,
$\displaystyle f'(x)=1+2x \sin(2/x)-2\cos(2/x)$ for $\displaystyle x\neq 0$.
To see if the derivative exists at $\displaystyle x=0$ just find the left and right limit of $\displaystyle f'(x)$ at $\displaystyle x=0$. The term $\displaystyle 2x \sin(2/x)$ is clearly 0 for both of these limits, but the term $\displaystyle 2\cos(2/x)$ flucuates between 1 and -1, so the limits do not exist and thus the derivitve does not exist at $\displaystyle x=0$.
I am still working on the second part but I am 100% sure you have typed the question incorrectly (a simple graph will show you it is always increasing). So you probably need to show f is not DECREASING, in which case you show the derivative is always positive.
I'm sorry but the derivative does exist at x= 0. What lancekam has shown is that the derivative is not continous at x= 0.
To find the derivative at x= 0, use the definition: the derivative at x= 0 is given by
$\displaystyle \lim_{h\rightarrow 0}\frac{f(h)- f(0)}{h}= \lim_{h\rightarrow 0}\frac{h+ h^2sin(2/h)}{h}$$\displaystyle = \lim_{h\rightarrow 0} 1+ h sin(2/h)$.
sin(2/h) lies between -1 and 1 for all non-zero h while the h multiplying it goes to 0: the limit of h sin(2/yh) is 0. The derivative at x= 0 is 1.