Let the function f be  given by  $$f(x) = \begin{cases}x + x^2 \sin(\frac{\pi}{x}); & x  \neq 0 \\0; & x = 0\end{cases}$$ 

then $$\left( f'(1) - f'(0) \right)$$ is 

$$f'\left(x\right)=1+2x\sin\ \frac{\pi}{x}+x^2\cos\ \frac{\pi}{x}\left(-\frac{\pi}{x^2}\ \right)$$ = $$f'\left(x\right)=1+2x\sin\ \frac{\pi}{x}-\pi\ \cos\ \frac{\pi}{x}$$

=> $$f'\left(1\right)=1+2\sin\ \pi-\pi\ \cos\ \pi=1+\pi\ $$

$$f'\left(0\right)=\lim_{h->0}\ \frac{\left(f\left(0+h\right)-f\left(0\right)\right)}{h}$$ = $$f'\left(0\right)=\lim_{h->0}\ \frac{f\left(h\right)-0}{h}$$ = $$f'\left(0\right)=\lim_{h->0}\ \frac{\left(h+h^2\sin\ \frac{\pi}{h}\right)}{h}$$

=$$f'\left(0\right)=\lim_{h->0}\ \left(1+h\sin\ \frac{\pi}{h}\right)=1+0\ =1$$

Hence $$f'\left(1\right)-f'\left(0\right)=1+\pi\ -1=\pi\ $$